【kuangbin带你飞】 MST专题

时间：2019-11-16 23:16:49 阅读：91 评论：0 收藏：0 [点我收藏+]

标签：eth reac split positive 结束 cer font ane hit

唉，被班级合唱和复变考试搞得心力交瘁。新算法学不进去，更新下吧

A - Til the Cows Come Home

Input

The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.

Output

The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.

Sample Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0

Sample Output

216
30

思路：把字母转换成节点编号套下板子就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 50500;


///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[50500],n,m,ans,eu,ev,cnt,t;

struct node{
    int u, v, w;
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void kruskal()
{
    sort(e, e+m, cmp);
    for(int i = 0; i < m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        /*if(++cnt == n-1) {
            break;
        }*/
    }
}

int main()
{
    while(scanf("%d",&n)) {
        if(!n) break;
        for(int i = 1; i <= 28; i++) {
            fa[i] = i;
        }
        int t = n-1;
        m = 0;
        ans = 0;
        while(t--) {
            char ch;
            int num;
            cin >> ch >> num;
            while(num--) {
                char vge;
                int cost;
                cin >> vge >> cost;
                e[m].u = ch-64;
                e[m].v = vge-64;
                e[m].w = cost;
                m++;
            }
        }
        /*for(int i = 0; i < m; i++) {
            printf("e[%d].u:%d e[%d].v:%d e[%d].w:%d\n",i,e[i].u,i,e[i].v,i,e[i].w);
        }*/
        kruskal();

        printf("%d\n",ans);
    }
    return 0;
}

B - Networking

You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area.
Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.

Input

The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line.
The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.

Output

For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.

Sample Input

Sample Output

0
17
16
26

思路：板子题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt;

struct node{
    int u, v, w;
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

bool unite(int r1, int r2)
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int n, int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        //printf("!!! i:%d n:%d m:%d\n",i,n,m);
        if(unite(e[i].u,e[i].v)) {
            cnt++;
            ans += e[i].w;
            //printf("ans:%d\n",ans);
        }
        if(cnt == n-1) {
            break;
        }
    }
}

int main()
{
    while(scanf("%d",&n) && n) {
        scanf("%d",&m);
        init(n);
        for(int i = 1; i <= m; i++) {
            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);
        }
        kruskal(n,m);
        printf("%d\n",ans);
    }
    return 0;
}

C - Building a Space Station

You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task.
The space station is made up with a number of units, called cells. All cells are sphere-shaped, but their sizes are not necessarily uniform. Each cell is fixed at its predetermined position shortly after the station is successfully put into its orbit. It is quite strange that two cells may be touching each other, or even may be overlapping. In an extreme case, a cell may be totally enclosing another one. I do not know how such arrangements are possible.

All the cells must be connected, since crew members should be able to walk from any cell to any other cell. They can walk from a cell A to another cell B, if, (1) A and B are touching each other or overlapping, (2) A and B are connected by a `corridor‘, or (3) there is a cell C such that walking from A to C, and also from B to C are both possible. Note that the condition (3) should be interpreted transitively.

You are expected to design a configuration, namely, which pairs of cells are to be connected with corridors. There is some freedom in the corridor configuration. For example, if there are three cells A, B and C, not touching nor overlapping each other, at least three plans are possible in order to connect all three cells. The first is to build corridors A-B and A-C, the second B-C and B-A, the third C-A and C-B. The cost of building a corridor is proportional to its length. Therefore, you should choose a plan with the shortest total length of the corridors.

You can ignore the width of a corridor. A corridor is built between points on two cells‘ surfaces. It can be made arbitrarily long, but of course the shortest one is chosen. Even if two corridors A-B and C-D intersect in space, they are not considered to form a connection path between (for example) A and C. In other words, you may consider that two corridors never intersect.

Input

The input consists of multiple data sets. Each data set is given in the following format.

n
x1 y1 z1 r1
x2 y2 z2 r2
...
xn yn zn rn

The first line of a data set contains an integer n, which is the number of cells. n is positive, and does not exceed 100.

The following n lines are descriptions of cells. Four values in a line are x-, y- and z-coordinates of the center, and radius (called r in the rest of the problem) of the sphere, in this order. Each value is given by a decimal fraction, with 3 digits after the decimal point. Values are separated by a space character.

Each of x, y, z and r is positive and is less than 100.0.

The end of the input is indicated by a line containing a zero.

Output

For each data set, the shortest total length of the corridors should be printed, each in a separate line. The printed values should have 3 digits after the decimal point. They may not have an error greater than 0.001.

Note that if no corridors are necessary, that is, if all the cells are connected without corridors, the shortest total length of the corridors is 0.000.

Sample Input

3
10.000 10.000 50.000 10.000
40.000 10.000 50.000 10.000
40.000 40.000 50.000 10.000
2
30.000 30.000 30.000 20.000
40.000 40.000 40.000 20.000
5
5.729 15.143 3.996 25.837
6.013 14.372 4.818 10.671
80.115 63.292 84.477 15.120
64.095 80.924 70.029 14.881
39.472 85.116 71.369 5.553
0

Sample Output

20.000
0.000
73.834

思路：把所有点先存起来，算出每个点之间的距离。
注意：POJ这个东西得double，G++和C++轮着交

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m;
double ans;
int eu,ev,cnt;

struct node {
    int u,v;
    double w;
}e[maxn];

struct Node {
    double x,y,z,r;
}a[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

bool unite(int r1, int r2)
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        //if(++cnt == n-1) {
            //break;
        //}
    }
}

int main()
{
    while(scanf("%d",&n) && n) {
        for(int i = 1; i <= n; i++) {
           scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);
        }
        //int u,v,w;
        int num = 1;
        for(int i = 1; i < n; i++) {
            for(int j = i+1; j <= n; j++) {
                e[num].u = i, e[num].v = j;
                double temp;
                temp = sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y)+(a[i].z-a[j].z)*(a[i].z-a[j].z))- a[i].r - a[j].r;
                temp = max(0.0,temp);
                e[num].w = temp;
                num++;

            }
        }
        num--;
        init(n);
        kruskal(num);
        printf("%.3lf\n",ans);
    }
    return 0;
}

D - Constructing Roads

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

Sample Output

179

思路：存点算间距，给出的已连通的直接让他们距离为0就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt;

struct node{
    int u, v, w;
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        if(++cnt == n-1) {
            break;
        }
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF) {
        int num = 1;
        int temp;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                cin >> temp;
                if(i == j) continue;
                e[num].u = i, e[num].v = j;
                e[num].w = temp;
                num++;

            }
        }
        num--;

        int q;
        cin >> q;
        while(q--) {
            int x,y;
            cin >> x >> y;
            for(int i = 1; i <= num; i++) {
                if(e[i].u == x && e[i].v == y) {
                    e[i].w = 0;
                }
                if(e[i].u == y && e[i].v == x) {
                    e[i].w = 0;
                }
            }
        }
        init(n);
        kruskal(num);
        printf("%d\n",ans);
    }
    return 0;
}

E - Truck History

Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of exactly seven lowercase letters (each letter on each position has a very special meaning but that is unimportant for this task). At the beginning of company‘s history, just a single truck type was used but later other types were derived from it, then from the new types another types were derived, and so on.

Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan -- i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as
1/Σ_{(t_o,t_d)}d(t_o,t_d)
where the sum goes over all pairs of types in the derivation plan such that t _o is the original type and t _d the type derived from it and d(t _o,t _d) is the distance of the types.
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan.

Input

The input consists of several test cases. Each test case begins with a line containing the number of truck types, N, 2 <= N <= 2 000. Each of the following N lines of input contains one truck type code (a string of seven lowercase letters). You may assume that the codes uniquely describe the trucks, i.e., no two of these N lines are the same. The input is terminated with zero at the place of number of truck types.

Output

For each test case, your program should output the text "The highest possible quality is 1/Q.", where 1/Q is the quality of the best derivation plan.

Sample Input

4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0

Sample Output

The highest possible quality is 1/3.

思路：题意有点难以理解，模拟下给出的公式就知道什么意思了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 6e6+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt;

struct node {
    int u, v, w;
}e[maxn];

struct Node {
    char s[10];
}a[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        if(++cnt == n-1) {
            break;
        }
    }
}

int main()
{
    while(scanf("%d",&n) && n) {
        for(int i = 1; i <= n; i++) {
            cin >> a[i].s;
            //printf("len:%d\n",a[i].s.length());
        }
        int num = 1;
        for(int i = 1; i < n; i++) {
            for(int j = i+1; j <= n; j++) {
                e[num].u = i, e[num].v = j;
                int temp = 0;
                for(int k = 0; k < 7; k++) {
                    if(a[i].s[k] != a[j].s[k])
                        temp++;
                }
                e[num].w = temp;
                num++;
            }
        }
        num--;
        init(n);
        kruskal(num);
        printf("The highest possible quality is 1/%d.\n",ans);
    }
    return 0;
}

F - Arctic Network

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel.
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

Input

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

Output

For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

Sample Input

Sample Output

212.13

思路：存点算距离

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt;

struct node{
    int u, v;
    double w;
}e[maxn];

struct Node {
    double x,y;
}a[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int n,int s,int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        //printf("cnt:%d n-s:%d\n",cnt+1,n-s);
        if(++cnt == n-s) {
            printf("%.2lf\n",e[i].w);
            break;
        }
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        int s,n;
        scanf("%d %d",&s,&n);
        for(int i = 1; i <= n; i++) {
            scanf("%lf %lf",&a[i].x,&a[i].y);
        }
        int num = 1;
        for(int i = 1; i <= n; i++) {
            for(int j = i+1; j <= n; j++) {
                e[num].u = i, e[num].v = j;
                e[num].w = sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));
                num++;
            }
        }
        /*for(int i = 1; i <= num; i++) {
            printf("e[%d]:%lf\n",i,e[i].w);
        }*/
        num--;
        init(n);
        kruskal(n,s,num);
    }
    return 0;
}

G - Highways

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can‘t reach via a highway. It is necessary to build more highways so that it will be possible to drive between any pair of towns without leaving the highway system.

Flatopian towns are numbered from 1 to N and town i has a position given by the Cartesian coordinates (xi, yi). Each highway connects exaclty two towns. All highways (both the original ones and the ones that are to be built) follow straight lines, and thus their length is equal to Cartesian distance between towns. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.

The Flatopian government wants to minimize the cost of building new highways. However, they want to guarantee that every town is highway-reachable from every other town. Since Flatopia is so flat, the cost of a highway is always proportional to its length. Thus, the least expensive highway system will be the one that minimizes the total highways length.

Input

The input consists of two parts. The first part describes all towns in the country, and the second part describes all of the highways that have already been built.

The first line of the input file contains a single integer N (1 <= N <= 750), representing the number of towns. The next N lines each contain two integers, xi and yi separated by a space. These values give the coordinates of i ^th town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location.

The next line contains a single integer M (0 <= M <= 1000), representing the number of existing highways. The next M lines each contain a pair of integers separated by a space. These two integers give a pair of town numbers which are already connected by a highway. Each pair of towns is connected by at most one highway.

Output

Write to the output a single line for each new highway that should be built in order to connect all towns with minimal possible total length of new highways. Each highway should be presented by printing town numbers that this highway connects, separated by a space.

If no new highways need to be built (all towns are already connected), then the output file should be created but it should be empty.

Sample Input

Sample Output

1 6
3 7
4 9
5 7
8 3

思路：多组输入wa，存点计算各点之间的距离，然后kruskal

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 1055;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt;

struct Node
{
    int x, y;
}a[maxn];

void init(int n)
{
    cnt = 0;
    ans = 0;
    for (int i = 0; i <= n; i++)
        fa[i] = i;
}

struct node
{
    int u, v, w;
    bool operator < (const node& A) const
    {
        return w < A.w;
    }
}edge[maxn * maxn];

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal()
{
    sort(edge, edge + m);
    for (int i = 0; i < m; i++)
    {
        int u = edge[i].u, v = edge[i].v, w = edge[i].w;
        if (unite(u, v))
        {
            printf("%d %d\n", u, v);
            cnt++;
            if (cnt == n -1 ) break;
        }
    }
}

int main()
{
    scanf("%d", &n);
    {
        m = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d", &a[i].x, &a[i].y);
            for (int j = 1; j < i; j++)
            {
                edge[m].u = i;
                edge[m].v = j;
                edge[m++].w = (a[i].x - a[j].x) * (a[i].x - a[j].x) +
                              (a[i].y - a[j].y) * (a[i].y - a[j].y);
            }
        }


        int Q;
        scanf("%d", &Q);
        while (Q--)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            if (unite(a, b)) cnt++;
        }
        kruskal();
    }
    return 0;
}

H - Agri-Net

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

Sample Output

28

思路：存点计算各点间距，然后kruskal

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 2e5+7;
typedef long long LL;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m;
LL ans;
int eu,ev,cnt;

struct node{
    int u, v, w;
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int n, int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        if(++cnt == n-1) {
            break;
        }
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF) {
        int num = 1;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                int temp;
                scanf("%d",&temp);
                if(i != j) {
                    e[num].w = temp;
                    e[num].u = i, e[num].v = j;
                    num++;
                }
            }
        }
        num--;
        init(n);
        kruskal(n,num);
        printf("%lld\n",ans);
    }
    return 0;
}

I - Borg Maze

The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated subspace network that insures each member is given constant supervision and guidance.

Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.

Input

On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character, a space `` ‘‘ stands for an open space, a hash mark ``#‘‘ stands for an obstructing wall, the capital letter ``A‘‘ stand for an alien, and the capital letter ``S‘‘ stands for the start of the search. The perimeter of the maze is always closed, i.e., there is no way to get out from the coordinate of the ``S‘‘. At most 100 aliens are present in the maze, and everyone is reachable.

Output

For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.

Sample Input

2
6 5
##### 
#A#A##
# # A#
#S  ##
##### 
7 7
#####  
#AAA###
#    A#
# S ###
#     #
#AAA###
#####

Sample Output

8
11

思路：对每个S或A进行编号，bfs求各点之间互相到达的最短距离建图，然后kruskal

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt,xx,yy;
char maze[60][60];
int a[60][60];
int dr[4] = {1, 0, -1, 0};
int dc[4] = {0, 1, 0, -1};
int num;
bool vis[60][60];

struct node {
    int u, v, w;
}e[maxn];

struct Node {
    int x, y;
    int step;
};

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal()
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        if(++cnt == n-1) {
            //break;
        }
    }
}

void bfs(int x, int y, int ans)
{
    memset(vis,0,sizeof(vis));
    queue <Node> q;
    Node temp;
    temp.x = x, temp.y = y, temp.step = 0;
    vis[x][y] = 1;
    q.push(temp);
    while(!q.empty()) {
        temp = q.front();
        q.pop();
        for(int i = 0; i < 4; i++) {
            Node nxt;
            nxt.x    = temp.x + dr[i];
            nxt.y    = temp.y + dc[i];
            nxt.step = temp.step + 1;
            //printf("a[%d][%d]:%d vis[%d][%d]:%d\n",nxt.x,nxt.y,a[nxt.x][nxt.y],nxt.x,nxt.y,vis[nxt.x][nxt.y]);
            if(nxt.x >= 1 && nxt.y >= 1 && nxt.x <= xx && nxt.y <= yy &&
                        !vis[nxt.x][nxt.y] && a[nxt.x][nxt.y] != -1) {
                vis[nxt.x][nxt.y] = 1;
                q.push(nxt);
                //printf("a[%d][%d] : %d\n",nxt.x,nxt.y,a[nxt.x][nxt.y]);
                if(a[nxt.x][nxt.y] > 0) {
                    e[m].u = ans, e[m].v = a[nxt.x][nxt.y];
                    e[m++].w = nxt.step;
                    //printf("u:%d v:%d w:%d\n",e[m-1].u,e[m-1].v,e[m-1].w);
                    //printf("num!:%d\n",m);
                }
            }
        }
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d %d",&yy, &xx);
        getchar();
        n = 1, m = 1;
        memset(a,0,sizeof(a));
        for(int i = 1; i <= xx; i++) {
            char ch[60];
            gets(ch);
            for(int j = 0; j < yy; j++) {
                if(ch[j] == ‘#‘) a[i][j+1] = -1;
                else if(ch[j] == ‘A‘ || ch[j] == ‘S‘) a[i][j+1] = n++;
                else if(ch[j] == ‘ ‘) a[i][j+1] = 0;
                //printf("a[%d][%d]:%d\n",i,j+1,a[i][j+1]);
            }
        }
        //printf("\n");
        for(int i = 1; i <= xx; i++) {
            for(int j = 1; j <= yy; j++) {
                if(a[i][j] > 0) {
                   bfs(i,j,a[i][j]);
                    //printf("num:%d\n",m);
                }
            }
        }
        //printf("n:%d m:%d\n",n,m);
        n--;
        m--;
        init(n);
        kruskal();
        printf("%d\n",ans);
    }
    return 0;
}

J - The Unique MST

Given a connected undirected graph, tell if its minimum spanning tree is unique.

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V‘, E‘), with the following properties:
1. V‘ = V.
2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E‘) of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E‘.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string ‘Not Unique!‘.

Sample Input

Sample Output

3
Not Unique!

思路：对于唯一生成树问题，我们可以通过对重边的标记，在进行kruskal时跳过用过的重边，判断是否能够找到边权相同的最小生成树。也有另外一种通过求次小生成树来判断MST是否唯一的方法，后续会补上。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt,ok,flag;

struct node{
    int u, v, w;
    int used,del,rep;//使用过，删掉惹，重复边
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
        e[i].del = 0, e[i].rep = 0, e[i].used = 0;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

int kruskal(int m)
{
    ans = 0;

    for(int i = 1; i <= m; i++) {
        //printf("!\n");
        if(e[i].del) continue;
        //printf("?\n");
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        //printf("eu:%d ev:%d\n",eu,ev);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        //cout << "e[i].w:" << e[i].w << endl;
        if(!flag) e[i].used = 1;
        fa[ev] = eu;
        //printf("ok:%d e[i].w:%d e[i-1].w:%d\n",ok,e[i].w,e[i-1].w);
        if(++cnt == n-1) {
            break;
        }
    }
    //printf("ans:%d\n",ans);
    return ans;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        ok = 0;
        scanf("%d %d",&n,&m);
        for(int i = 1; i <=m; i++) {
            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);
        }
        init(n);
        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= m; j++) {
                if(j == i) continue;
                if(e[i].w == e[j].w) {
                    e[i].rep = 1;
                    //printf("e[%d]:%d\n",i,e[i].rep);
                }
            }
        }
        flag = 0;
        sort(e+1, e+m+1, cmp);
        int res = kruskal(m);
        flag = 1;
        for(int i = 1; i <= n; i++) {
            //printf("e[%d]:%d\n",i,e[i].rep);
            if(e[i].rep == 1 && e[i].used == 1) {
                //init(n);
                e[i].del = 1;
                int temp;
                for(int g = 1; g <= n ; g++) {
                    fa[g] = g;
                }
                temp = kruskal(m);
                e[i].del = 0;
                //printf("temp:%d\n",temp);
                if(temp == res) {
                    printf("Not Unique!\n");
                    ok = 1;
                    break;
                }
            }
        }
        if(!ok) {
            printf("%d\n",res);
        }
    }
    return 0;
}

K - 还是畅通工程

某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。

Input测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。
当N为0时，输入结束，该用例不被处理。
Output对每个测试用例，在1行里输出最小的公路总长度。
Sample Input

Sample Output

3
5


       
 
Huge input, scanf is recommended.

Hint

Hint
        

思路：建图+kruskal

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,ans,eu,ev,cnt;

struct node{
    int u, v, w;
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0;
    cnt = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        if(++cnt == n-1) {
            break;
        }
    }
}

int main()
{
    m = 1;
    while(scanf("%d",&n) && n) {
        int m = n * (n-1) / 2;
        for(int i = 1; i <= m; i++) {
            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);
        }
        init(n);
        kruskal(m);
        printf("%d\n",ans);
    }

    return 0;
}

L - Jungle Roads

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.

The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.

The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.

Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0

Output

216
30

Sample Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0

Sample Output

216
30

思路：别问，问就是bin巨的锅

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 50500;

int fa[50500],n,m,ans,eu,ev,cnt,t;

struct node{
    int u, v, w;
}e[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void kruskal()
{
    sort(e, e+m, cmp);
    for(int i = 0; i < m; i++) {
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        ans += e[i].w;
        fa[ev] = eu;
        /*if(++cnt == n-1) {
            break;
        }*/
    }
}

int main()
{
    while(scanf("%d",&n)) {
        if(!n) break;
        for(int i = 1; i <= 28; i++) {
            fa[i] = i;
        }
        int t = n-1;
        m = 0;
        ans = 0;
        while(t--) {
            char ch;
            int num;
            cin >> ch >> num;
            while(num--) {
                char vge;
                int cost;
                cin >> vge >> cost;
                e[m].u = ch-64;
                e[m].v = vge-64;
                e[m].w = cost;
                m++;
            }
        }
        /*for(int i = 0; i < m; i++) {
            printf("e[%d].u:%d e[%d].v:%d e[%d].w:%d\n",i,e[i].u,i,e[i].v,i,e[i].w);
        }*/
        kruskal();

        printf("%d\n",ans);
    }
    return 0;
}

M - 畅通工程再续

相信大家都听说一个“百岛湖”的地方吧，百岛湖的居民生活在不同的小岛中，当他们想去其他的小岛时都要通过划小船来实现。现在政府决定大力发展百岛湖，发展首先要解决的问题当然是交通问题，政府决定实现百岛湖的全畅通！经过考察小组RPRush对百岛湖的情况充分了解后，决定在符合条件的小岛间建上桥，所谓符合条件，就是2个小岛之间的距离不能小于10米，也不能大于1000米。当然，为了节省资金，只要求实现任意2个小岛之间有路通即可。其中桥的价格为 100元/米。

Input输入包括多组数据。输入首先包括一个整数T(T <= 200)，代表有T组数据。
每组数据首先是一个整数C(C <= 100),代表小岛的个数，接下来是C组坐标，代表每个小岛的坐标，这些坐标都是 0 <= x, y <= 1000的整数。
Output每组输入数据输出一行，代表建桥的最小花费，结果保留一位小数。如果无法实现工程以达到全部畅通，输出”oh!”.Sample Input

Sample Output

1414.2

思路：建图后kruskal，注意一些细节，比如double和对t比到1e(-6)级，另外一些题意上的坑点详见hdu里此题讨论区

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const int maxn = 2e5+7;

///把所有边排序，记第i小的边为:e[i](1 <= i < m)
///初始化MST为空
///初始化连通分量，让每个点自成一个独立的连通分量
///for(int i = 0; i < m; i++) {
///   if(e[i].u 和 e[i].v 不在同一个连通分量) {
///     把边e[i]加入MST
///     合并e[i].u 和 e[i].v 所在的连通分量
///   }
///}

int fa[5050],n,m,eu,ev,cnt,ok;
double ans;

struct node{
    int u, v;
    double w;
}e[maxn];

struct Node {
    int x,y;
}a[maxn];

bool cmp(node a, node b)
{
    return a.w < b.w;
}

int fid(int x)
{
    return x == fa[x] ? x : fid(fa[x]);
}

void init(int n)
{
    for(int i = 1; i <= n; i++) {
        fa[i] = i;
    }
    ans = 0.0;
    cnt = 0;
    ok  = 0;
}

bool unite(int r1, int r2)///冰茶鸡
{
    int fidroot1 = fid(r1), fidroot2 = fid(r2);
    if(fidroot1 != fidroot2) {
        fa[fidroot2] = fidroot1;
        return true;
    }
    return false;
}

void kruskal(int m)
{
    sort(e+1, e+m+1, cmp);
    for(int i = 1; i <= m; i++) {
        //printf("e[%d]:%f\n",i,e[i].w);
        eu = fid(e[i].u);
        ev = fid(e[i].v);
        if(eu == ev) {
            continue;
        }
        if(e[i].w >= 10.000000 && e[i].w <= 1000.0000001) {
            ans += e[i].w;
            fa[ev] = eu;
            //printf("ans:%.1f cnt:%d\n",ans,cnt);

            if(++cnt == n-1) {
                ok = 1;
                break;
            }
        }
    }
    if(ok) {
        printf("%.1f\n",ans * 100);
    }
    else {
        printf("oh!\n");
    }
}

double dist(Node a, Node b)
{
    return sqrt((a.x - b.x)*(a.x - b.x)+(a.y - b.y)*(a.y - b.y));
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(int i = 1; i <= n; i++) {
            scanf("%d %d",&a[i].x,&a[i].y);
        }
        m = 1;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                if(i == j) {
                    continue;
                }
                else {
                    e[m].u = i, e[m].v = j;
                    e[m++].w = dist(a[i],a[j]);
                }
            }
        }
        m--;
        init(n);
        kruskal(m);
    }
    return 0;
}

【kuangbin带你飞】 MST专题

标签：eth reac split positive 结束 cer font ane hit

原文地址：https://www.cnblogs.com/orangeko/p/11874425.html

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