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Connections between cities LCA

时间:2018-08-09 11:10:05      阅读:174      评论:0      收藏:0      [点我收藏+]

标签:produced   owb   stdin   build   ***   one   题意   span   uri   

Problem Description
After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.
 

 

Input
Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.
 

 

Output
For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.
 

 

Sample Input
5 3 2 1 3 2 2 4 3 5 2 3 1 4 4 5
 

 

Sample Output
Not connected 6
Hint
Hint Huge input, scanf recommended.

 

 

题意是说给你一个森林,让你求两点之间的最近距离。
lca求最近公共祖先,如果不是在同一棵树上,则输出Not connected。

用并查集来判断是否在同一颗树上面

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <set>
  7 #include <iostream>
  8 #include <map>
  9 #include <stack>
 10 #include <string>
 11 #include <vector>
 12 #define  pi acos(-1.0)
 13 #define  eps 1e-6
 14 #define  fi first
 15 #define  se second
 16 #define  lson l,m,rt<<1
 17 #define  rson m+1,r,rt<<1|1
 18 #define  bug         printf("******\n")
 19 #define  mem(a,b)    memset(a,b,sizeof(a))
 20 #define  fuck(x)     cout<<"["<<x<<"]"<<endl
 21 #define  f(a)        a*a
 22 #define  sf(n)       scanf("%d", &n)
 23 #define  sff(a,b)    scanf("%d %d", &a, &b)
 24 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
 25 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
 26 #define  pf          printf
 27 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
 28 #define  FREE(i,a,b) for(i = a; i >= b; i--)
 29 #define  FRL(i,a,b)  for(i = a; i < b; i++)
 30 #define  FRLL(i,a,b) for(i = a; i > b; i--)
 31 #define  FIN         freopen("DATA.txt","r",stdin)
 32 #define  gcd(a,b)    __gcd(a,b)
 33 #define  lowbit(x)   x&-x
 34 #pragma  comment (linker,"/STACK:102400000,102400000")
 35 using namespace std;
 36 typedef long long LL;
 37 typedef unsigned long long ULL;
 38 const int maxn = 1e5 + 10;
 39 int _pow[maxn], dep[maxn], dis[maxn], vis[maxn], ver[maxn];
 40 int tot, head[maxn], dp[maxn * 2][25], k, first[maxn], fa[maxn];
 41 struct node {
 42     int u, v, w, nxt;
 43 } edge[maxn << 2];
 44 void init() {
 45     tot = 0;
 46     mem(head, -1);
 47     for (int i = 0 ; i < maxn ; i++) fa[i] = i;
 48 }
 49 int Find(int x) {
 50     return x == fa[x] ? fa[x] : fa[x] = Find(fa[x]);
 51 }
 52 void combine(int x, int y) {
 53     int nx = Find(x), ny = Find(y);
 54     if(nx != ny) fa[nx] = ny;
 55     return ;
 56 }
 57 void add(int u, int v, int w) {
 58     edge[tot].v = v, edge[tot].u = u;
 59     edge[tot].w = w, edge[tot].nxt = head[u];
 60     head[u] = tot++;
 61 }
 62 void dfs(int u, int DEP) {
 63     vis[u] = 1;
 64     ver[++k] = u;
 65     first[u] = k;
 66     dep[k] = DEP;
 67     for (int i = head[u]; ~i; i = edge[i].nxt) {
 68         if (vis[edge[i].v]) continue;
 69         int v = edge[i].v, w = edge[i].w;
 70         dis[v] = dis[u] + w;
 71         dfs(v, DEP + 1);
 72         ver[++k] = u;
 73         dep[k] = DEP;
 74     }
 75 }
 76 void ST(int len) {
 77     int K = (int)(log((double)len) / log(2.0));
 78     for (int i = 1 ; i <= len ; i++) dp[i][0] = i;
 79     for (int j = 1 ; j <= K ; j++) {
 80         for (int i = 1 ; i + _pow[j] - 1 <= len ; i++) {
 81             int a = dp[i][j - 1], b = dp[i + _pow[j - 1]][j - 1];
 82             if (dep[a] < dep[b]) dp[i][j] = a;
 83             else dp[i][j] = b;
 84         }
 85     }
 86 }
 87 int RMQ(int x, int y) {
 88     int K = (int)(log((double)(y - x + 1)) / log(2.0));
 89     int a = dp[x][K], b = dp[y - _pow[K] + 1][K];
 90     if (dep[a] < dep[b]) return a;
 91     else return b;
 92 }
 93 int LCA(int u, int v) {
 94     int x = first[u], y = first[v];
 95     if (x > y) swap(x, y);
 96     int ret = RMQ(x, y);
 97     return ver[ret];
 98 }
 99 int main() {
100     for (int i = 0 ; i < 40 ; i++) _pow[i] = (1 << i);
101     int n, m, q;
102     while(~sfff(n, m, q)) {
103         init();
104         mem(vis, 0);
105         for (int i = 0 ; i < m ; i++) {
106             int u, v, w;
107             sfff(u, v, w);
108             add(u, v, w);
109             add(v, u, w);
110             combine(u, v);
111         }
112         k = 0;
113         for (int i = 1 ; i <= n ; i++) {
114             if (fa[i] == i) {
115                 dis[i] = 0;
116                 dfs(i, 1);
117             }
118         }
119         ST(2 * n - 1);
120         while(q--) {
121             int u, v;
122             sff(u, v);
123             int lca = LCA(u, v);
124             if (Find(u) == Find(v)) printf("%d\n", dis[u] + dis[v] - 2 * dis[lca]);
125             else printf("Not connected\n");
126         }
127     }
128     return  0;
129 }

 

Connections between cities LCA

标签:produced   owb   stdin   build   ***   one   题意   span   uri   

原文地址:https://www.cnblogs.com/qldabiaoge/p/9447264.html

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