View Code of Problem 128

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<climits>
#include<cmath>
#include<map>
#include<set>

using namespace std;


//bool dfs(int time, int pos, int n) {
//
//	if (pos == 1 && time < 0)
//		return true;
//	if (pos == 1 && time >= 0)
//		return false;
//
//	for (int i = n - 1; i >= 1; i--) {
//
//		if (city[i][pos] != 0) {	//存在路径
//
//			if (dfs(time + city[i][pos], i, n))
//				return true;
//		}
//	}
//
//	return false;
//}


int main()
{
	int m, n;

	while (cin >> n >> m) {

		vector<int> dp(n + 1, INT_MAX);	//从城市1到其他城市的最短距离
		dp[1] = 0;

		vector<vector<int>> city(n + 1, vector<int>(n + 1, 0));

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

			int a, b, t;
			cin >> a >> b >> t;

			city[a][b] = t;
		}


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

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

				if (city[i][j] != 0) {

					dp[j] = min(dp[j], dp[i] + city[i][j]);
				}
			}
		}

		int flag = 0;
		for (int i = 1; i <= n; i++) {

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

				if (city[i][j] != 0) {

					if (dp[i] + city[i][j] < dp[j]) {

						flag = 1;
						break;	//存在负权回路
					}
				}
			}
		}

		
		if(dp[n] < 0 || flag == 1)
			cout << "YES" << endl;
		else if(dp[n] >= 0)
			cout << "NO" << endl;

		//if (dfs(0, n, n))
		//	cout << "YES" << endl;
		//else
		//	cout << "NO" << endl;
	}
}