学步园

题意：出于竞争目的，股票经纪人要在他们的人际关系网中传播一些谣言。但是每个股票经纪人只能传他给他所熟悉的几个人。当谣言从第一个人传出后，最快经过多多长时间最后一个人可以收到谣言。
题解：通过Floyd求出任意两个人之间的cost。然后分析一下即可。

#include <iostream>
using namespace std;

#define INF 200000000
int contract[101][101], n;

void Floyd ()
{
	int i,j,k;
	for ( k = 1; k <= n; k++ )
	{
		for ( i = 1; i <= n; i++ )
			for ( j = 1; j <= n; j++ )
				if ( contract[i][j] > contract[i][k] + contract[k][j] )
					contract[i][j] = contract[i][k] + contract[k][j];
	}
}


int main()
{
	int i, j, k, t;
	while ( cin >> n && n )
	{
		for ( i = 1; i <= n; i++ )
			for ( j = 1; j <= n; j++ )
				contract[i][j] = ( i == j ? 0 : INF );

		for ( i = 1; i <= n; i++ )
		{
			cin >> k;
			for ( j = 1; j <= k; j++ )
			{
				cin >> t;
				cin >> contract[i][t];
			}
		}
			
		Floyd ();
		int start = -1, temp, time = INF;
		for ( i = 1; i <= n; i++ )
		{
			temp = 0;
			for ( j = 1; j <= n; j++ )
			{
				if ( contract[i][j] == INF )
					break;
				if ( contract[i][j] > temp )
					temp = contract[i][j];
			}

			if ( temp < time && j == n+1 )
			{
				time = temp;
				start = i;
			}
		}

		if ( start == -1 )
			cout << "disjoint" << endl;
		else
			cout << start << ' ' << time << endl;
	}
	return 0;
}