最短路(Floyd_Warshall) POJ 1125 Stockbroker Grapevine

题目传送门

 /*
     最短路：Floyd模板题
             主要是两点最短的距离和起始位置
     http://blog.****.net/y990041769/article/details/37955253
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <vector>
 using namespace std;

 const int MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 int d[MAXN][MAXN];

 void Floyd_Warshall(int n)
 {
     for (int k=; k<=n; ++k)
     {
         for (int i=; i<=n; ++i)
         {
             for (int j=; j<=n; ++j)
             {
                 if (d[i][j] > d[i][k] + d[k][j])
                     d[i][j] = d[i][k] + d[k][j];
             }
         }
     }
 }

 void work(int n)
 {
     Floyd_Warshall (n);
     int res;   int ans = INF;  int num;
     for (int i=; i<=n; ++i)
     {
         res = ;
         for (int j=; j<=n; ++j)
         {
             if (i == j) continue;
             if (res < d[i][j])  res = d[i][j];
         }
         if (ans > res)
         {
             ans = res;
             num = i;
         }
     }

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

 int main(void)      //POJ 1125  Stockbroker Grapevine
 {
     //freopen ("E.in", "r", stdin);

     int n, m;
     while (~scanf ("%d", &n) && n)
     {
         for (int i=; i<=n; ++i)
             for (int j=; j<=n; ++j)    d[i][j] = INF;
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &m);
             int x, y;
             for (int j=; j<=m; ++j)
             {
                 scanf ("%d%d", &x, &y);
                 d[i][x] = y;
             }
         }

         work (n);
     }

     return ;
 }