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。

结题思路：

　　　　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|-12     For each edge(u,v)属于G.E3         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\n");73         else printf("YES\n");74     }75     return 0;76 }