poj3259 Bellman_Ford算法

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.

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 1 #include<stdio.h>
 2 #include<string.h>
 3
 4 const int INF=99999999;
 5
 6 struct Node{
 7     int u;
 8     int v;
 9     int o;
10 };
11
12 Node  edge[10250];
13 int dis[550];
14 int n,m,w,nn;
15
16 void relax(int u,int v,int o)
17 {
18     if(dis[v]>dis[u]+o)
19         dis[v]=dis[u]+o;
20 }
21
22 bool Bellan_Ford()
23 {
24     int i,j,k;
25     for(i=1;i<=n;i++)
26         dis[i]=INF;
27
28     for(i=1;i<=n-1;i++)
29     {
30         for(j=1;j<=nn;j++)
31             relax(edge[j].u,edge[j].v,edge[j].o);
32     }
33
34     for(i=1;i<=nn;i++)
35     {
36         if(dis[edge[i].v]>dis[edge[i].u]+edge[i].o)
37         {
38             return false;
39         }
40     }
41     return true;
42 }
43
44 int main()
45 {
46     int F,i,j,k,flg;
47     int s,e,t;
48     scanf("%d",&F);
49     while(F--)
50     {
51         flg=0;nn=1;
52         scanf("%d %d %d",&n,&m,&w);
53         for(i=1;i<=m;i++)
54         {
55             scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o);
56             nn++;
57             edge[nn].u=edge[nn-1].v,edge[nn].v=edge[nn-1].u,edge[nn].o=edge[nn-1].o;
58             nn++;
59         }
60         for(i=1;i<=w;i++)
61         {
62             scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o);
63             edge[nn].o=-edge[nn].o;
64             nn++;
65         }
66         nn--;
67         if(Bellan_Ford()==false)
68         {
69             flg=1;
70         }
71         if(flg==1)
72             printf("YES\n");
73         else
74             printf("NO\n");
75     }
76     return 0;
77 }