poj3259Wormholes【SPFA判负环】

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Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 38297 Accepted: 14093

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int maxn=5100;struct Node{int from,to,next,value;}A[maxn];int vis[510],dist[510];int head[510],inqueue[510];int node;void add(int s,int t,int c){A[node].from=s;A[node].to=t;A[node].value=c;A[node].next=head[s];head[s]=node++;}bool SPFA(int n){queue<int>Q;memset(vis,0,sizeof(vis));memset(inqueue,0,sizeof(inqueue));memset(dist,0x3f,sizeof(dist));Q.push(1);vis[1]++;dist[1]=0;while(!Q.empty()){int top=Q.front();Q.pop();inqueue[top]=0;for(int k=head[top];k!=-1;k=A[k].next){if(dist[A[k].to]>dist[top]+A[k].value){dist[A[k].to]=dist[top]+A[k].value;if(!inqueue[A[k].to]){inqueue[A[k].to]=1;vis[A[k].to]++;if(vis[A[k].to]>n-1)return true;Q.push(A[k].to);}}}}return false;}int main(){int t,i,j,n,m,w;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&w);memset(head,-1,sizeof(head));int a,b,c;node=0;for(i=0;i<m;++i){scanf("%d%d%d",&a,&b,&c);add(a,b,c);add(b,a,c);}for(i=0;i<w;++i){scanf("%d%d%d",&a,&b,&c);add(a,b,-c);}if(SPFA(n)){printf("YES\n");}else {printf("NO\n");}}return 0;}