POJ - 3259 Wormholes ~~spfa判断是否有负环

Wormholes

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

就是让你看下~~主角是否能够通过穿越单向的虫洞（权值为负）来达到不断返回过去的情况~~建成图之后就是判定是否存在负环的存在问题~~

#include<cstdio>
#include<queue>
#include<iostream>
using namespace std;
int m, n;
int head[1000];//链式前向星
struct edge
{
	int to;
	int ne ;
	int len;
}ed[10000];//储存边~~
int time[1000];//用来判断是否有负环
bool vis[1000];
int d[1000];//距离（在这里代表时间）
int cnt = 0;
void init()//初始化
{
	for (int s = 0; s <= 999; s++)
	{
		head[s] = -1;
	}
	for (int s = 0; s <= n; s++)
	{
		d[s] = 999999;
	}
	for (int s = 0; s < 9000; s++)
	{
		ed[s].ne = -1;
	}
	memset(vis, 0, sizeof(vis));
	memset(time, 0, sizeof(time));
}
void add(int from, int to, int len)
{
	ed[cnt].to = to;
	ed[cnt].len = len;
	ed[cnt].ne = head[from];
	head[from] = cnt++;
}
bool spfa()//返回是否能有负环
{
	d[1] = 0;
	queue<int>q;
	q.push(1);
	vis[1] = 1;
	bool flag = false;
	while (!q.empty())
	{
		int t = q.front();
		q.pop();
	//	cout << t << endl;
	//	cout << head[t] << endl;
		for (int s = head[t]; ~s; s = ed[s].ne)
		{
		//	cout << t << " " << ed[s].to << endl;
			if (d[ed[s].to] > d[t] + ed[s].len)
			{
				d[ed[s].to] = d[t] + ed[s].len;
				if (!vis[ed[s].to])
				{
			//		cout << ed[s].to << endl;
					vis[ed[s].to] = 1;
					time[ed[s].to]++;
					if (time[ed[s].to] > n)
					{
						flag = true;
						return flag;
					}
					q.push(ed[s].to);
				}
			}
		}
		vis[t] = 0;
	}
	return flag;
}
int main()
{
	int w, te;
	scanf("%d", &te);
	while (te--)
	{
		cin >> n >> m >> w;
		cnt = 0;
		init();
		for (int s = 0; s < m; s++)
		{
			int a, b, c;
			cin >> a >> b >> c;
			add(a, b, c);//开始的时候是双向边
			add(b, a, c);
		}
		for (int s = 0; s < w; s++)
		{
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			add(a, b, -c);//虫洞~~负边
		}
	//	cout << " " << d[3] << endl;
		bool ans = spfa();
		if (ans)
		{
			cout << "YES" << endl;
		}
		else
		{
			cout << "NO" << endl;
		}
		while (1)
		{
			break;
		}
	//	cout << d[n] << endl;
	}
	return 0;
}