Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Description

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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

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.

Source


SPFA找负环。
AC代码:
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
struct node
{
	int v, w, next;
}e[5500];
int k, n, m, w;
int head[5500];
bool vis[510];
int dis[510];
queue <int> q;
void adde(int u, int v, int w)
{
	e[k].v = v;
	e[k].w = w;
	e[k].next = head[u];
	head[u] = k++;
}
void input()
{
	cin >> n >> m >> w;
	int s, e, t;
	k = 1;
	memset(head, -1, sizeof(head));
	for (int i = 0; i < m; i++)
	{
		cin >> s >> e >> t;
		adde(s, e, t);
		adde(e, s, t);
	}
	for (int i = 0; i < w; i++)
	{
		cin >> s >> e >> t;
		adde(s, e, -t);
	}
}
bool spfa(int x)
{
	memset(vis, 0, sizeof(vis));
	while (!q.empty())
		q.pop();
	memset(dis, 0x3f, sizeof(dis));
	dis[x] = 0;
	vis[x] = 1;
	q.push(x);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		vis[u] = 0;
		for (int i = head[u]; i != -1; i = e[i].next)
		{
			int &v = e[i].v;
			if (dis[v] > dis[u] + e[i].w)
			{
				dis[v] = dis[u] + e[i].w;
				if (!vis[v])
				{
					vis[v] = 1;
					q.push(v);
				}
			}
		}
		if (dis[x] < 0)
			return 1;
	}
	return 0;
}
void solve()
{
	for (int i = 1; i <= n; i++)
	{
		if (spfa(i))
		{
			cout << "YES\n";
			return;
		}
	}
	cout << "NO\n";
}
int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		input();
		solve();	
	}
	return 0;
}

 

分类: ACM/ICPC

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