poj_1125 Floyd最短路

题目大意

题目分析

    典型的最短路问题，由于最开始选择的出发点不固定，因此需要求出所有点之间的最短路。采用floyd算法。判断图是否连通，可以通过判断是否从图中所有的点出发都存在无法到达的点来实现：若从图中所有的点出发，都存在无法到达的点，则说明图不连通。

实现(c++)

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
#define INFINITE 1 << 28
int gDist[105][105];

//Floyd算法
void Floyd(int n){
	for (int k = 1; k <= n; k++){ //从i到j，中间经过的节点编号不大于k
		for (int i = 1; i <= n; i++){ //起点i
			for (int j = 1; j <= n; j++){ //终点j
				if (gDist[i][j] > gDist[i][k] + gDist[k][j]){
					gDist[i][j] = gDist[i][k] + gDist[k][j];
				}
			}
		}
	}
}
int main(){
	int n;
	while (scanf("%d", &n) && n){
		int k, v, d;
		for (int i = 1; i <= n; i++){
			for (int j = 1; j <= n; j++){
				gDist[i][j] = INFINITE;
				if (i == j)
					gDist[i][j] = 0;
			}
		}
		for (int i = 1; i <= n; i++){
			scanf("%d", &k);
			for (int j = 0; j < k; j++){
				scanf("%d %d", &v, &d);
				gDist[i][v] = d;
			}
		}
		Floyd(n);
		int min_time = INFINITE, min_stockbroker = 0, disjoint_count = 0;
		
		for (int i = 1; i <= n; i++){
			int max = 0;
			bool disjoint = false;
			for (int j = 1; j <= n; j++){
				if (gDist[i][j] == INFINITE){   //说明存在点i无法到达的点
					disjoint = true;
					break;
				}
				max = max > gDist[i][j] ? max : gDist[i][j];
			}
			if (disjoint){
				disjoint++;
				continue;
			}
			if (min_time > max){
				min_time = max;
				min_stockbroker = i;
			}
			
		}
		if (disjoint_count == n)    //如果从每个点出发都有无法到达的点，则说明
		                            //图不是连通的
			printf("disjoint\n");
		else
			printf("%d %d\n", min_stockbroker, min_time);
	}
	return 0;
}