Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34302		Accepted: 12520

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.

 

 

 

以每个点为起点跑一次，检测是否有负环。

1 #include 
       
         2 #include 
        
          3 using    namespace    std; 4  5 const    int    INF = 0xfffffff; 6 const    int    SIZE = 5500; 7 int    D[505]; 8 int    N,M,W,F; 9 struct    Node10 {11     int    from,to,cost;12 }G[SIZE];13 14 bool    Bellman_Ford(int);15 bool    relax(int,int,int);16 int    main(void)17 {18     scanf("%d",&F);19     while(F --)20     {21         scanf("%d%d%d",&N,&M,&W);22         int    i = 0;23         while(i < 2 * M)24         {25             scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);26             i ++;27             G[i] = G[i - 1];28             swap(G[i].from,G[i].to);29             i ++;30         }31         while(i < W + M * 2)32         {33             scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);34             G[i].cost = -G[i].cost;35             i ++;36         }37         bool    flag = true;38         for(int i = 1;i <= N;i ++)39             if(!Bellman_Ford(i))40             {41                 puts("YES");42                 flag = false;43                 break;44             }45         if(flag)46             puts("NO");47     }48 49     return    0;50 }51 52 bool    Bellman_Ford(int s)53 {54     fill(D,D + 505,INF);55     D[s] = 0;56     bool    update;57 58     for(int i = 0;i < N - 1;i ++)59     {60         update = false;61         for(int i = 0;i < M * 2 + W;i ++)62             if(relax(G[i].from,G[i].to,G[i].cost))63                 update = true;64         if(!update)65             break;66     }67     for(int i = 0;i < M * 2 + W;i ++)68         if(relax(G[i].from,G[i].to,G[i].cost))69             return    false;70     return    true;71 }72 73 bool    relax(int from,int to,int cost)74 {75     if(D[to] > D[from] + cost)76     {77         D[to] = D[from] + cost;78         return    true;79     }80     return    false;81 }