Wormholes
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
题目大意：
FJ童鞋家的农田里面有若干个虫洞，有一些虫洞是相连的，可以消耗一些时间爬过去(无向)。还有一些虫洞比较特殊，从一头爬到那一头之后时间会退后一些(有向)
FJ希望通过爬虫洞(?)来时时间退后，从而能看到另外一个自己。根据输入的虫洞信息来判断是否可以实现FJ的愿望。
解题思路：
说白了就是判断有向图有没有负环。Bellman-Ford算法和SPFA算法都可以用来判断是否有负环。
1.Bellman-Ford算法：
1 For i=1 to |G.V|-1 2 For each edge(u,v)属于G.E 3 RELAX(u,v,w) 4 For each edge(u,v)属于G.E //此循环用来判断负环 5 If (v.d>u.d+w(u,v) 6 Return FALSE; 7 Return TRUE;
2.SPFA算法：
可以通过判断 顶点i进入队列的次数是否大于N-1来判断是否存在负环。
PS：相对于没有负环的有N个顶点的有向图来说，一个顶点最多松弛N-1次即可达到最短路。
Code(SPFA):
1 #include<stdio.h> 2 #include<string> 3 #include<iostream> 4 #include<limits.h> 5 #include<queue> 6 using namespace std; 7 int edge[600][600],times[5505],dis[5505]; 8 bool vis[5505]; 9 int N; 10 void init(int N) 11 { 12 for (int i=1; i<=N; i++) 13 for (int j=1; j<=N; j++) 14 edge[i][j]=INT_MAX; 15 } 16 bool SPFA(int begin) 17 { 18 for (int i=1; i<=N; i++) 19 { 20 dis[i]=INT_MAX; 21 vis[i]=0; 22 times[i]=0; 23 } 24 queue<int> Q; 25 Q.push(begin); 26 dis[begin]=0; 27 vis[begin]=1; 28 times[begin]++; 29 while (!Q.empty()) 30 { 31 begin=Q.front(); 32 Q.pop(); 33 vis[begin]=0; 34 for (int j=1; j<=N; j++) 35 if (j!=begin&&edge[begin][j]!=INT_MAX&&dis[j]>dis[begin]+edge[begin][j]) 36 { 37 dis[j]=edge[begin][j]+dis[begin]; 38 if (!vis[j]) 39 { 40 Q.push(j); 41 vis[j]=1; 42 times[j]++; 43 if (times[j]>=N) return 0; 44 } 45 } 46 } 47 return 1; 48 } 49 int main() 50 { 51 int T; 52 cin>>T; 53 while (T--) 54 { 55 int M,W; 56 cin>>N>>M>>W; 57 init(N); 58 for (int i=1; i<=M; i++) 59 { 60 int x1,x2,x3; 61 scanf("%d%d%d",&x1,&x2,&x3); 62 if (x3<edge[x1][x2]) 63 edge[x1][x2]=edge[x2][x1]=x3; 64 } 65 for (int i=1; i<=W; i++) 66 { 67 int x1,x2,x3; 68 scanf("%d%d%d",&x1,&x2,&x3); 69 edge[x1][x2]=0-x3; 70 } 71 bool ok=SPFA(1); 72 if (ok) printf("NO "); 73 else printf("YES "); 74 } 75 return 0; 76 }