POJ 3615 Cow Hurdles

http://poj.org/problem?id=3615

floyd 最短路径的变形 dist[i][j]变化为 : i j之间的最大边

那么输入的时候可以直接把dist[i][j] 当作i j 之间的边进行输入

转移方程 dist[i][j] = max(dist[i][j], min(dist[i][k], dist[k][j]))

 #include <iostream>
 #include <string.h>
 #include <stdio.h>
 #define INF 0x3fffffff
 using namespace std;
 const int maxn = ;
 int Map[maxn][maxn];

 int dist[maxn][maxn];//dist[i][j]表示 i 到 j 的最大边是多少？
 //转移方程 dist[i][j] = min( dist[i][j] ,max(dist[i][k], dist[k][j])) ---> "**"
 //
 //与floyd的区别 dist是 i-j的最短'距离'
 int main()
 {
     int m, n, k;
     scanf("%d%d%d", &m, &n, &k);
     for(int i = ; i <= m; i++)
         for (int j = ; j <= m; j++)
             dist[i][j] = INF;
     for (int i = ; i < n; i++)
     {
         int from, to, cost;
         scanf("%d%d%d", &from, &to, &cost);
         dist[from][to] = cost;
     }
     for(int i = ; i <= m; i++)
     {
         for (int j = ; j <= m; j++)
         {
             for (int k = ; k <= m; k++)
             {
                // if (dist[j][k] == -1) dist[j][k] = max(dist[j][i], dist[i][k]);
                 //else
                     dist[j][k] = min(dist[j][k], max(dist[j][i], dist[i][k]));
             }
         }
     }
     for (int i = ; i < k; i++)
     {
         int from, to;
         scanf("%d%d", &from, &to);
         if (dist[from][to] == INF) cout << - << endl;
         else  cout << dist[from][to] << endl;
     }
     return ;
 }