POJ 1125 Stockbroker Grapevine

给你n个人的联系情况，对任意一个人，求出这个人发消息到其他n-1个人的时间，得到n-1个时间中的最大值，n个最大值中的最小值就是所求。如果网络不通，那就输出disjoint

Floyed算出任意两个人的最小时间 就OK了

#include <iostream>
#include <fstream>
using namespace std;
const int MAXN = 101;
const int INF = 0x7FFF;
int m , n;
int G[MAXN][MAXN];

void in(){
    int i,j,w,t;
    for(i = 1; i <= m;i++)
        for(j = 1; j <= m; j++)
            G[i][j] = INF;
    for(i = 1; i <= m; i++)
        G[i][i] = 0;
    for(i = 1; i <= m; i++){
        cin >> n;
        for(j = 1; j <= n;j++){
            cin >> w >> t;
            G[i][w] = t;
        }
    }
}

void floyd(){
    int i,j,k;
    for(k = 1; k <= m; k++)
        for(i = 1; i <= m; i++)
            for(j = 1; j <= m; j++){
                if(G[i][j] > G[i][k] + G[k][j])
                    G[i][j] = G[i][k] + G[k][j];
            }

}

void out(){
    int i,j,temp = 0,bestN,best = INF;
    bool flag = false;
    for(i = 1; i <= m; i++){
        flag = false;
        temp = 0;
        for(j = 1; j <= m; j++){
            if( temp < G[i][j] && i != j){
                temp = G[i][j];
            }
            if(G[i][j] == INF){
                flag = true;
                break;
            }
        }
        if(!flag){
            if(best > temp ){
                best = temp;
                bestN = i;
            }      
        }
    }
    if(bestN == INF)
        cout << "disjoint" << endl;
    else
        cout << bestN << " " << best << endl;
}

int main(){
    while(cin >> m, m!=0){
        in();
        floyd();
        out();
    }
    return 0;
}