【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

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
");
70             else printf("NO
");
71         }
72         return 0;
73     }