Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37906		Accepted: 13954

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.

正常的path是双向的，有一定的消耗时间。虫洞是单向的，能够让时间倒流一定时间。问FJ能否找到一条路径，能够遇见之前的那个自己。

说白了就是找负环。

Bellman_ford模板题，用来对每一条边都进行松弛，然后看最后结果是否依然能够松弛。如果还能松弛，说明有负环；如果不能松弛了，就是没有负环。

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #pragma warning(disable:4996)using namespace std;struct E{ int s; int e; int l;}edge[5205];int N,M,W,edge_num;int dis[505];void addedge(int start,int end,int len){ edge_num++; edge[edge_num].s=start; edge[edge_num].e=end; edge[edge_num].l=len;}bool bellman_ford(){ int i,j; for(i=1;i<=N-1;i++) { int flag=0; for(j=1;j<=edge_num;j++) { if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l) { flag=1; dis[edge[j].e]=dis[edge[j].s]+edge[j].l; } } if(flag==0) break; } for(j=1;j<=edge_num;j++) { if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l) { return true; } } return false;}int main(){ int i,start,end,len; int Test; cin>>Test; while(Test--) { edge_num=0; memset(dis,0,sizeof(dis)); cin>>N>>M>>W; for(i=1;i<=M;i++) { cin>>start>>end>>len; addedge(start,end,len); addedge(end,start,len); } for(i=1;i<=W;i++) { cin>>start>>end>>len; addedge(start,end,-len); } if(bellman_ford()) { cout<<"YES"<
         
        
       
      
     
    

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