POJ 3259 Wormholes

题目链接 ：http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 35789 Accepted: 13074

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.

思路：判断是否有负圈，若有，则能够达到目的；

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cstdlib>using namespace std;#define man 10050struct node{int v;int w;int next; } edg[man];  int n,m,w; int head[man]; int cnn; int visit[man]; int dis[man]; int vis[man]; #define INF 0x7fffff void add(int u ,int v,int wi) { edg[cnn].v=v; edg[cnn].w=wi; edg[cnn].next=head[u]; head[u]=cnn++; } void spfa() { memset(visit,0,sizeof(visit)); memset(vis,0,sizeof(vis)); queue<int>q; fill(dis,dis+n+1,INF); visit[1]=1; dis[1]=0; q.push(1); vis[1]++; while(!q.empty()) { int u=q.front(); q.pop(); visit[u]=0; int i; for(i=head[u];i!=-1;i=edg[i].next) { int v=edg[i].v; if(dis[v]>dis[u]+edg[i].w) { dis[v]=dis[u]+edg[i].w; if(!visit[v]) { visit[v]=1; vis[v]++; if(vis[v]>=n){ cout<<"YES"<<endl; return; }  q.push(v); }  } } } cout<<"NO"<<endl; } int main(){ int t; cin >> t; while(t--) { cin >>n >> m >> w; memset(head,-1,sizeof(head)); cnn=0; for(int i=1;i<=m;i++) { int  a,b,c; cin>>a>>b>>c; add(a,b,c); add(b,a,c); } for(int i=1;i<=w;i++) { int d,f,g; cin >> d >> f >>g; add(d,f,-g); } spfa();  } }