hdu 5137 去掉一个点 使得最短路最大（2014广州现场赛 K题）

题意：从2~n-1这几个点中任意去掉一个点，使得从1到n的最短路径最大，如果任意去掉一个点1~n无通路输出Inf。

Sample Input
4 5
1 2 3
1 3 7
1 4 50
2 3 4
3 4 2
3 2
1 2 30
2 3 10
0 0

Sample Output
50
Inf

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <string>
# include <map>
# include <cmath>
# include <queue>
# include <list>
# define LL long long
using namespace std ;

const int MAXN=300;
const int INF=0x3f3f3f3f;
int n  , m ;
bool vis[MAXN];
int cost[MAXN][MAXN] ;
int lowcost[MAXN] ;

void Dijkstra(int beg)
{
    for(int i=0;i<n;i++)
    {
        lowcost[i]=INF;
    }
    lowcost[beg]=0;
    for(int j=0;j<n;j++)
    {
        int k=-1;
        int Min=INF;
        for(int i=0;i<n;i++)
            if(!vis[i]&&lowcost[i]<Min)
            {
                Min=lowcost[i];
                k=i;
            }
            if(k==-1)
                break ;
            vis[k]=true;
            for(int i=0;i<n;i++)
                if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i])
                {
                    lowcost[i]=lowcost[k]+cost[k][i];
                }
    }

}

int main ()
{
    //freopen("in.txt","r",stdin) ;
    while (scanf("%d %d" , &n , &m) !=EOF)
    {
        if (n == 0 && m == 0)
            break ;

        int u , v , w ;
        int i , j ;
        int MAX = 0 ;
        for (i = 0 ; i < n ; i++)
        for (j = 0 ; j < n ; j++)
       {
           if (i == j)
              cost[i][j] = 0 ;
           else
              cost[i][j] = INF ;
       }
        while(m--)
        {
            scanf("%d%d%d" , &u , &v , &w) ;
            cost[u-1][v-1] = w ;
            cost[v-1][u-1] = w ;

        }

        for (i = 1 ; i <= n-2 ; i++)
        {
            memset(vis , 0 , sizeof(vis)) ;
            vis[i] = 1 ;
            Dijkstra(0) ;
            if (lowcost[n-1] > MAX )
                MAX = lowcost[n-1] ;

        }

        if (MAX == INF)
           printf("Inf
") ;
        else
           printf("%d
" , MAX) ;
    }

    return 0 ;
}