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

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

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

         }

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

         }

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

     return  ;
 }