【最短路】poj3259-Wormholes（Bellman-Ford 最短路）

                                                                                           Wormholes
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 32109   Accepted: 11660

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
 
题意： John的农场里N块地，M条路连接两块地，W个虫洞，虫洞是一条单向路，会在你离开之前把你传送到目的地，就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。简化下，就是看图中有没有负权环。有的话就是可以，没有的话就是不可以了。
 
       这道题的题意不好懂，主要难理解的是：
        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.   一个是双向边
        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.  一个是单向边，而且单向边为负值。
 
 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 
 5 const int VM=520;
 6 const int EM=5020;
 7 const int INF=0x3f3f3f3f;
 8 
 9 struct Edge{
10     int u,v;
11     int c;
12 }edge[EM<<1];
13 
14 int n, m, w, cnt, dis[VM];
15 
16 
17     void add(int u, int v, int c)
18     {
19         edge[cnt].u = u;
20         edge[cnt].v = v;
21         edge[cnt].c = c;
22         cnt++;
23     }
24     int Bellman()
25     {
26         int i, j, k, flag;
27         for(i=0; i<n; i++)
28             dis[i] = INF;
29         for(i=1; i<n; i++)
30         {
31             flag = 0;
32             for(j=0; j<cnt; j++)
33             {
34                 if(dis[edge[j].v] > dis[edge[j].u] + edge[j].c)
35                 {
36                     dis[edge[j].v] = dis[edge[j].u] + edge[j].c;
37                     flag = 1;
38                 }
39             }
40             if(!flag) break;
41         }
42         for(k=0; k<cnt; k++)
43             if(dis[edge[k].v] > dis[edge[k].u] + edge[k].c)
44                 return 1;
45         return 0;
46 
47     }
48     int main()
49     {
50         int T, i;
51         scanf("%d", &T);
52         while(T--)
53         {
54             int x, y, z;
55             cnt = 0;
56             scanf("%d%d%d", &n, &m, &w);
57             for(i=0; i<m; i++)
58             {
59                 scanf("%d%d%d", &x, &y, &z);
60                 add(x, y, z);//加边
61                 add(y, x, z);
62             }
63             for(i=0; i<w; i++)
64             {
65                 scanf("%d%d%d", &x, &y, &z);
66                 add(x, y, -z);//注意这里是-z
67             }
68             if(Bellman())
69                 printf("YES\n");
70             else printf("NO\n");
71         }
72         return 0;
73     }

 

 
 
  
posted @ 2014-12-02 20:51  6bing  阅读(218)  评论(0)  编辑  收藏  举报