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|-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\n");
73 else printf("YES\n");
74 }
75 return 0;
76 }
POJ3259——Wormholes(Bellman-Ford+SPFA)