Bellman_ford POJ 3259 Wormholes

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 41728		Accepted: 15325

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

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 could start from anywhere on the cycle to
accomplish this.

 

分析：John的农场里N块地，M条路连接两块地，W个虫洞；路是双向的，虫洞是一条单向路，会在你离开之前把你传送到目的地，
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
简化下，就是看图中有没有负权环。

 

 #include <cstdio>
 #define maxn 502
 #define INF 0x3fffffff

 ///边
 typedef struct Edge{
     int u, v;    ///起点， 终点
     int weight;  ///权值
 }Edge;

 Edge edge[];     ///双向边，保存边的权值
 int  dist[maxn];     ///节点到原点的最小距离
 int edgenum;    ///边数
 ///插入边
 void insert(int u, int v, int w)
 {
     edge[edgenum].u = u;
     edge[edgenum].v = v;
     edge[edgenum++].weight = w;
 }
 bool Bellman_Ford(int source, int nodenum)  ///原点和结点个数
 {
     for(int i=; i < nodenum; ++i)
         dist[i] = INF;
     dist[source] = ;
     for(int i=; i < nodenum; ++i){
         for(int j=; j < edgenum; ++j)
         {
             if(dist[edge[j].v] > dist[edge[j].u] + edge[j].weight)///松弛计算
                 dist[edge[j].v] = dist[edge[j].u] + edge[j].weight;
         }
     }
     bool flag = false;
     /// 判断是否有负权环
     for(int i=; i < edgenum; ++i){
         if(dist[edge[i].v] > dist[edge[i].u] + edge[i].weight)
         {
             flag = true;  ///有负权环
             break;
         }
     }
     return flag;
 }
 int main()
 {
     int T;
     scanf("%d", &T);
     while(T--)
     {
         int n, m, c;
         scanf("%d%d%d", &n, &m, &c);
         edgenum = ;
         for(int i=; i<m; i++)
         {
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             insert(u, v, w);
             insert(v, u, w);
         }
         for(int i=; i<c; i++)
         {
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             insert(u, v, -w);
         }
         if(Bellman_Ford(, n))
             printf("YES\n");
         else
             printf("NO\n");
     }
     return ;
 }