POJ 1125 Stockbroker Grapevine

解题思路: 首先Floyd算法求出所有顶点之间的最小距离，计算最小最大距离

#include <iostream>
using namespace std;

#define MaxNum 1e8

int main()
{
    int p[105][105];
    int dist[105];
    int n, m, v, t, start, minTime;
    while (cin >> n && n)
    {
        
        fill(&p[0][0], &p[104][104], MaxNum);
        for (int i = 1; i <= n; i++)
        {
            p[i][i] = 0;
            cin >> m;
            while(m--)
            {
                cin >> v >> t;
                p[i][v] = t;
            }
        }

        for (int k = 1; k <= n; k++)
            for (int i = 0; i <= n; i++)
                for (int j = 0; j <= n; j++)
                    if (p[i][j] > p[i][k] + p[k][j])
                        p[i][j] = p[i][k] + p[k][j];

        minTime = MaxNum;
        for (int i = 1; i <= n; i++)
        {
            int maxTime = 0;
            for (int j = 1; j <= n; j++)
                if (maxTime < p[i][j])
                    maxTime = p[i][j];
            if(minTime > maxTime)
            {
                minTime = maxTime;
                start = i;
            }
        }
        if (minTime == MaxNum)
            cout << "disjoint"<<endl;
        else
            cout << start<< " " << minTime << endl;
    }
    return 0;
}