Buy or Build (poj 2784 最小生成树)

标签：buy or build   poj 2784   最小生成树   

Description

Input

Output

Sample Input

7 3
2 4 1 2
3 3 3 6 7
3 9 2 4 5
0 2
4 0
2 0
4 2
1 3
0 5
4 4

Sample Output

17

Hint

Sample Explanation: The above instance is shown in Figure 2. An optimal solution is described in Figure 3 (thick edges come from an existing network while thin edges have been setup from scratch).


Figure 3: An optimal solution of the 7 City instance in which which the first and second existing networkshave been bought while two extra edges (1, 5) and (2, 4)

Source

题意：n个城市，告诉每个城市的坐标，还有q个联通块，现在要把这n个城市连起来，可以购买联通块（每个有一定的费用），或者新建一条边（费用为点之间的距离的平方），问最小费用是多少。

思路：q很小，二进制枚举选哪些块，每次kruskal一遍，求最小值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

const int maxn = 1005;
const int MAXN = 500500;

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

struct Edge
{
    int u,v,len;
    bool operator<(const Edge &a)const
    {
        return len<a.len;
    }
}edge[MAXN];

int father[maxn],cost[10];
int n,q,num;
vector<int>g[10];

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

void addedge(int u,int v)
{
    edge[num].u=u;
    edge[num].v=v;
    edge[num++].len=(node[u].x-node[v].x)*(node[u].x-node[v].x)+(node[u].y-node[v].y)*(node[u].y-node[v].y);
}

int find_father(int x)
{
    if (x!=father[x])
        father[x]=find_father(father[x]);
    return father[x];
}

bool Union(int a,int b)
{
    int fa=find_father(a);
    int fb=find_father(b);
    if (fa==fb) return false;
    father[fa]=fb;
    return true;
}

int Kruskal()
{
    int ans=0;
    int cnt=0;
    for (int i=0;i<num;i++)
    {
        if (Union(edge[i].u,edge[i].v))
        {
            ans+=edge[i].len;
            cnt++;
        }
        if (cnt==n-1) break;
    }
    return ans;
}

void solve()
{
    init();
    int ans=Kruskal();
    for (int i=0;i<(1<<q);i++)
    {
        init();
        int all=0;
        for (int j=0;j<q;j++)
        {
            if (!((i>>j)&1)) continue;
            all+=cost[j];
            for (int k=1;k<g[j].size();k++)
                Union(g[j][k],g[j][0]);
        }
        ans=min(ans,all+Kruskal());
    }
    pf("%d\n",ans);
}

int main()
{
    int i,j,t,number,x;
//    sf(t);
//    while (t--)
    {
        sff(n,q);
        num=0;
        for (i=0;i<q;i++)
        {
            g[i].clear();
            sff(number,cost[i]);
            for (j=0;j<number;j++)
            {
                sf(x);
                g[i].push_back(x);
            }
        }
        for (i=1;i<=n;i++)
            sff(node[i].x,node[i].y);
        for (i=1;i<=n;i++)
            for (j=i+1;j<=n;j++)
                addedge(i,j);
        sort(edge,edge+num);
        solve();
//        if (t) puts("");
    }
    return 0;
}
/*
1

7 3
2 4 1 2
3 3 3 6 7
3 9 2 4 5
0 2
4 0
2 0
4 2
1 3
0 5
4 4
*/

Buy or Build (poj 2784 最小生成树)

标签：buy or build   poj 2784   最小生成树   

原文地址：http://blog.csdn.net/u014422052/article/details/46827821