poj 1125 Stockbroker Grapevine

题意：以任意顶点为起点，能够连通所有的节点，问你以哪个节点为起点时到所有的节点的最短路的最长路最短；

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<cstring>
#include<vector>
using namespace std;
const int inf = 0x7fffffff;
int map[124][124],N; 
int Dij( int start )
{
    int dis[124],hash[124];
    for( int i = 0 ; i <= N ; i ++ )
    {
        dis[i] = inf ; hash[i] = 0;
    }    
    dis[start] = 0;
    for( int i = 1 ; i <= N ; i ++ )
    {
       int min = inf ,t;
       for( int j = 1 ; j <= N ; j ++ )
       {
           if( !hash[j] && min > dis[j] )
           {
               min = dis[j] ; t = j;
            }        
       }    
       hash[t] = 1;
       for( int j = 1 ; j<= N ; j ++ )
       {
            if( !hash[j] && map[t][j] != inf && dis[j] > map[t][j] + dis[t] )
                dis[j] = dis[t] + map[t][j];        
       }
    }
    int ans = -1;
    for( int i = 1 ; i <= N ; i++ )
    {
        if( dis[i] > ans  ) 
        {
            ans = dis[i];
    //    printf( "ans=%d\n",ans );
       }    
    }
    return ans;
}
void empty(  )
{
     for( int i =  0; i <= N; i ++ )
     {
        for( int j = 0 ; j <= N ; j ++ )
           map[i][j] = inf;        
     }    
}
int main(  )
{
    int n,num,T;
    while( scanf( "%d",&N ),N )
    {
       empty();
       for( int i = 1 ; i <= N ; i ++ )
       {
          scanf( "%d",&n );
          for( int j = 0 ; j < n  ; j++ )
          {
              scanf( "%d %d",&num,&T );
              map[i][num] = T;
          }        
       }
       int t = 0 , time = inf;    
       for( int i = 1 ; i <= N ; i ++ )
       {
          t = Dij( i  );
          if( time > t)
          {
             num = i ; time = t;        
          }
       }
       if( time !=inf )
       printf( "%d %d\n",num , time );
       else printf( "disjoint\n" );
    }
    //system( "pause" );
    return 0;
}