POJ--3529--Wormholes

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36889		Accepted: 13520

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

题意：就是给N组測试数据，给你N个点M条无向边，再给你W个虫洞（能够使时间倒流）。问你能不能看到刚出发时的自己。（就是推断存不存在负环）。

思路：套上SPFA公式然后推断负环就能够了。

#include<stdio.h>//好坑，wa了半天没过去，竟然是由于大写和小写问题。整个人都不好了。
#include<string.h>
#include<queue>
#define INF 0x3f3f3f3f
#define M 3000*2
#define N 550
using namespace std;
int n,m,w,dis[N],vis[N],used[N],head[N],edgenum;
struct node{
	int from,to,cost,next;
}edge[M];
void init(){
	edgenum=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v,int cost){
	node E={u,v,cost,head[u]};
	edge[edgenum]=E;
	head[u]=edgenum++;
}
void spfa(){
	queue<int>q;
	memset(vis,0,sizeof(vis));
	memset(dis,INF,sizeof(dis));
	memset(used,0,sizeof(used));
	dis[1]=0;
	vis[1]=1;
	q.push(1);
	used[1]++;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=head[u];i!=-1;i=edge[i].next){
			int v=edge[i].to;
			if(dis[v]>dis[u]+edge[i].cost){
				dis[v]=dis[u]+edge[i].cost;
				if(!vis[v]){
					vis[v]=1;
					used[v]++;
					if(used[v]>n){
						printf("YES\n");//存在负环就意味着时光能够倒流。
						return ;
					}
					q.push(v);
				}
			}
		}
	}
	printf("NO\n");
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d%d",&n,&m,&w);
		init();
		while(m--){
			int a,b,cost;
			scanf("%d%d%d",&a,&b,&cost);//加入无向边。
			add(a,b,cost);
			add(b,a,cost);
		}
		while(w--){
			int a,b,cost;//加入单向虫洞。

scanf("%d%d%d",&a,&b,&cost);
			cost=-cost;
			add(a,b,cost);
		}
		spfa();
	}
	return 0;
}

ac代码：