[USACO07NOV]牛栏Cow Hurdles

题目大意:有一个有向图,有多个询问,每个询问给你a和b,要你求出a->b路径上最小的最大值

题解:因为是求点对之间的最小的最大值,可以用变形的Floyd来解,f[i][j]=min(f[i][j],max(f[i][k],f[k][j]))

 

C++ Code:

 

#include<cstdio>
using namespace std;
const int maxn=310;
const int inf=0x7fffffff;
int n,m,t;
int f[maxn][maxn];
int max(int a,int b){return a>b?a:b;}
int main(){
    scanf("%d%d%d",&n,&m,&t);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)if (i^j)f[i][j]=inf;
    for (int i=0;i<m;i++){
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        f[a][b]=c;
    }
    for (int k=1;k<=n;k++)
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++)if (f[i][j]>max(f[i][k],f[k][j]))f[i][j]=max(f[i][k],f[k][j]);
    for (int i=0;i<t;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",(f[a][b]^inf)?f[a][b]:-1);
    }
    return 0;
}

 

posted @ 2017-11-24 19:41  Memory_of_winter  阅读(195)  评论(0编辑  收藏  举报