题目链接：558 - Wormholes

题目大意：给出n和m，表示有n个点，然后给出m条边，然后判断给出的有向图中是否存在负环。

解题思路：利用Bellman Ford算法，若进行第n次松弛时，还能更新点的权值，则说明有负环的存在。

#include <stdio.h>

#include <string.h>

#define min(a,b) (a)<(b)?(a):(b)

const int N = 10005;

const int INF = 0x3f3f3f3f;

int n, m, flag, d[N];

int x[N], y[N], v[N];

void init() {

	scanf("%d%d", &n, &m);

	for (int i = 0; i < m; i++)

		scanf("%d%d%d", &x[i], &y[i], &v[i]);

}

void BF() {

	flag = 0;

	for (int i = 0; i < n; i++) d[i] = INF;

	d[0] = 0;

	for (int k = 1; k < n; k++) {

		for (int i = 0; i < m; i++) {

			int a = x[i], b = y[i];

			if (d[a] < INF)

				d[b] = min(d[a] + v[i], d[b]);

		}

	}

	for (int i = 0; i < m; i++) {

		int a = x[i], b = y[i];

		if (d[b] > d[a] + v[i]) {

			flag = 1;

			return;

		}

	}

}

int main () {

	int cas;

	scanf("%d", &cas);

	while (cas--) {

		init();

		BF();

		printf("%s\n", flag ? "possible" : "not possible");

	}

	return 0;

}