H - Wormholes POJ - 3259(spfa判负环模板）

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.

题目大意： N个点，M条路 权值为正，W条路权值为负 直接用spfa判负环。
spfa判负环有两种方法
如果没有负环的图，从起点到终点最多也就N条路
1.直接判断一个点进入队列次数大于>N 即有负环 （耗时较长）
2.判断这条路更新次数大于N次 num[i]=num[x]+1,即为从x到i这条路走的总次数 耗时较短

这里附上一个大佬讲解spfa的连接
https://blog.youkuaiyun.com/forever_dreams/article/details/81161527

这里贴下 第二种代码:
其实这里存边用vector要更快一点…（不想改了）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<queue>
#define fi first
#define se second
#define FOR(a) for(int i=0;i<a;i++)
#define show(a) cout<<a<<endl;
#define show2(a,b) cout<<a<<" "<<b<<endl;
#define show3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl;
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef pair<P, int> LP;
const ll inf = 0x3f3f3f3f;
const int N = 1e5 + 10;
const ll mod = 998244353;


int n, m, t, x, y, k, sum, cnt, ans;
int r, c,vis[1000],dist[1000],num[1000];

int mp[1000][1000];

int spfa( )
{
	queue<int> q;
	for(int i=1;i<=n;i++) dist[i]=inf,num[i]=0,vis[i]=0;
	num[1]=1;
	q.push(1);
	dist[1]=0;
	vis[1]=1;
	while(!q.empty())
	{
		int x=q.front();
		//show(x)
		q.pop();
		vis[x]=0;
		for(int i=1;i<=n;i++)
		{
			//show2(dist[i],i);
			if(dist[i]>mp[x][i]+dist[x])
			{
				dist[i]=mp[x][i]+dist[x];
				num[i]=num[x]+1;
				if(num[i]>n) return 1;

				if(!vis[i])
				{
					q.push(i);
					vis[i]=1;
				}
			}
		}
	}
	return 0;
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{

		scanf("%d%d%d",&n,&m,&k);
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
		{
			if(i!=j) mp[i][j]=inf;
			else mp[i][j]=0;
		}
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d%d",&x,&y,&c);
			mp[x][y]=min(mp[x][y],c);
			mp[y][x]=min(mp[y][x],c);
		}
		for(int i=1;i<=k;i++)
		{
			scanf("%d%d%d",&x,&y,&c);
			mp[x][y]=-c;
		}
		int flag=0;

		if(spfa())
		{
			puts("YES");
			flag=1;
		}

		if(!flag) puts("NO");
	}


}