POJ 3615 Cow Hurdles //Floyd算法

题目描述

POJ3615 Cow Hurdles

解题思路

核心代码: f[i][j] = min(f[i][j],max(f[i][k],f[k][j]));

参考代码

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#define inf 0x3f3f3f3f#define MAX_V 310using namespace std;int f[MAX_V][MAX_V];int main(){    int N,M,T,u,v,h;    scanf("%d %d %d",&N,&M,&T);    memset(f,inf,sizeof(f));    for(int i = 1;i <= M;i++){        scanf("%d %d %d",&u,&v,&h);        f[u][v] = h;    }    for(int k = 1;k <= N;k++)        for(int i = 1;i <= N;i++)            for(int j = 1;j <= N;j++)                f[i][j] = min(f[i][j],max(f[i][k],f[k][j]));    for (int i = 0;i < T;i++){        scanf("%d %d",&u,&v);        if (f[u][v] == inf) printf("-1\n");        else    printf("%d\n",f[u][v]);    }    return 0;}