poj3268 Silver Cow Party（两次dijkstra）

https://vjudge.net/problem/POJ-3268

一开始floyd超时了。。

对正图定点求最短，对逆图定点求最短，得到任意点到定点的往返最短路。

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<stack>
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 #define INF 0x3f3f3f3f
 typedef unsigned long long ll;
 using namespace std;
 int t[][], rt[][];
 int d[], rd[], vis[];
 int n, m, x, a, b, T;
 void dijkstra(int dist[], int a[][])
 {
     memset(vis, , sizeof(vis));
     for(int i = ; i <= n; i++){
         dist[i] = INF;
     }
     dist[x] = ;
     for(int i = ; i <= n; i++){
         int mini = INF, k = -;
         for(int j = ; j <= n; j++){
             if(!vis[j]&&mini > dist[j]){
                 mini = dist[j];
                 k = j;
             }
         }
         vis[k] = ;
         for(int j = ; j <= n; j++){
             if(!vis[j]&&dist[j] > dist[k]+a[k][j]){
                 dist[j] = dist[k]+a[k][j];
             }
         }
     }
 }
 int main()
 {
     cin >> n >> m >> x;
     for(int i = ; i <= n; i++){
         for(int j = ; j <= n; j++){
             t[i][j] = INF; rt[i][j] = INF;
         }
     }
     for(int i = ; i <= m; i++){
         cin >> a >> b >> T;
         t[a][b] = T;//图
         rt[b][a] = T;//逆图
     }
     dijkstra(d, t);
     dijkstra(rd, rt);
     int maxm = -INF;
     for(int i = ; i <= n; i++){
         if(d[i]!=INF&&rd[i]!=INF&&d[i]+rd[i] > maxm){
             maxm = d[i]+rd[i];
         }
     }
     cout << maxm << endl;
     return ;
 }