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；

View Code

#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;

}