【poj3259】Wormholes 【USACO 2006 December Gold】

Description

While exploring his many farms, Farmer John hasdiscovered a number of amazing wormholes. A wormhole is very peculiar becauseit is a one-way path that delivers you to its destination at a time that isBEFORE 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 dothe following: start at some field, travel through some paths and wormholes,and return to the starting field a time before his initial departure. Perhapshe will be able to meet himself :) .

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

Input

Line 1: A single integer, F. F farm descriptionsfollow.
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) thatdescribe, respectively: a bidirectional path between S and E that requires Tseconds 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) thatdescribe, respectively: A one way path from S to E that also moves the travelerback T seconds.

Output

Lines 1..F: For each farm, output "YES" ifFJ can achieve his goal, otherwise output "NO" (do not include thequotes).

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 couldstart from anywhere on the cycle to accomplish this.

这道题本质上就是判断一个图中是否存在负环，所以只要跑一遍SPFA，判断是否有节点入队达到N次，如果达到N次，则有负环，下面是代码：

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<iostream>#include<queue>using namespace std;int vis[505],dis[505];struct line{int v,w;line *next;};line *head[505];line *make(){return (line *)(malloc(sizeof(line)));}void add(int u,int v,int w){line *x=make();x->v=v;x->w=w;x->next=head[u];head[u]=x;}void work(){memset(dis,0x7f,sizeof(dis));memset(vis,0,sizeof(vis));memset(head,0,sizeof(head));queue<int>dl;int n,m,w,i,u,v,t;scanf("%d%d%d",&n,&m,&w);for(i=0;i<m;i++){scanf("%d%d%d",&u,&v,&t);add(u,v,t);add(v,u,t);}for(i=0;i<w;i++){scanf("%d%d%d",&u,&v,&t);add(u,v,-1*t);}dl.push(1);dis[1]=0;while(!dl.empty()){int x=dl.front();dl.pop();if(vis[x]>=n){printf("YES\n");return;}for(line *now=head[x];now;now=now->next){if(dis[now->v]>dis[x]+now->w){dis[now->v]=dis[x]+now->w;dl.push(now->v);vis[now->v]++;}}}for(i=1;i<=n;i++){line *now=head[i],*x;while(now){x=now;now=now->next;free(x);}}printf("NO\n");}int main(){int t;scanf("%d",&t);while(t--){work();}return 0;}