Poj3259--Wormholes（Spfa 判负环）

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

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.

Source

USACO 2006 December Gold

RE: 农场分为几块， 有的地方存在虫洞, (有负权值（边权值为时间）的边)。 问能不能回到过去。  Spfa 判 负环。

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 using namespace std;
 6 const int INF = 0x3f3f3f3f;
 7 int dis[550], vis[550], used[550];   //used[]数组判负环；
 8 int n, m, t;
 9 struct Edge
10 {
11     int from, to, w, next;
12 } edge[5050];
13 int head[550], cnt;
14 void Add(int a, int b, int w)
15 {
16     Edge E = {a, b, w, head[a]};
17     edge[cnt] = E;
18     head[a] = cnt++;
19 }
20 bool Spfa(int src)
21 {
22      queue<int> q;
23     for(int i = 1; i <= n; i++)
24         dis[i] = INF;
25     dis[src] = 0; vis[src] = 1;
26     q.push(src);
27     while(!q.empty())
28     {
29     //    printf("1\n");
30         int u = q.front();
31         q.pop();
32         vis[u] = 0;
33         for(int i = head[u]; i != -1; i = edge[i].next)
34         {
35             int v = edge[i].to;
36             if(dis[v] > dis[u] + edge[i].w)
37             {
38                 dis[v] = dis[u] + edge[i].w;
39                 if(!vis[v])
40                 {
41                     vis[v] = 1;
42                     q.push(v);
43                     used[v]++;
44                 }
45             }
46             if(used[v] > n)     //存在负环；
47                 return true;
48         }
49     }
50     return false;
51 }
52 int main()
53 {
54     int nt;
55     scanf("%d", &nt);
56     while(nt--)
57     {
58         scanf("%d %d %d", &n, &m, &t);
59         int u, v, w; cnt = 0;
60         memset(vis, 0, sizeof(vis));
61         memset(used, 0 , sizeof(used));
62         memset(head, -1, sizeof(head));
63         for(int i = 0; i < m; i++)
64         {
65             scanf("%d %d %d", &u, &v, &w);
66             Add(u, v, w);
67             Add(v, u, w);
68         }
69         for(int i = 1; i <= t; i++)
70         {
71             scanf("%d %d %d", &u, &v, &w);
72             Add(u, v, -w);
73         }
74         if(Spfa(1))
75             printf("YES\n");
76         else
77             printf("NO\n");
78     }
79     return 0;
80 }