POJ 1125 Stockbroker Grapevine

/*
*POJ 1125 Stockbroker Grapevine
*maxdist 为从第i个人出发 传递完成所需的最大距离
*ans 为maxdist 中的最小值
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAXN = 110;
const int INF = 0x3f3f3f3f;
int dist[MAXN][MAXN];
int n;
int floyd()
{
	int i, j, k;
	for (k = 0; k < n; k ++) {
		for (i = 0;  i < n; i++) {
			for (j = 0; j < n; j++) {
				dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
			}
		}
	}
}

int main()
{
	int i, j, num;
	while (~scanf("%d", &n) && n) {
		memset(dist, INF, sizeof(dist));
		for (i = 0; i < n; i ++) {
			scanf("%d", &num);
			while (num--) {
				scanf("%d", &j);
				scanf("%d", &dist[i][j - 1]);
			}
		}
		floyd();
		int ans = INF, maxdist, pos;
		for (i = 0; i < n; i++) {
			maxdist = -1;
			for (j = 0; j < n; j ++) {
				if (i != j && maxdist < dist[i][j]) {
					maxdist = dist[i][j];
				}
			}
			if (ans > maxdist) {
				ans = maxdist;
				pos = i;
			}
		}
		if (ans == INF) {
			printf("disjoint\n");
		} else {
			printf("%d %d\n", pos+1, ans);
		}
	}
	return 0;
}