poj Wormholes 【最短路径】【bellman

poj Wormholes 【最短路径】【bellman_ford】

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

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.

【题意】有f组数据，n个场所，m条路，w个虫洞，经过虫洞连接的路后可以时光倒流。求是否可以从任意点出发后回到原点

【思路】flody算法只能用于权值是正数的最短路径，这里的虫洞的权值为负，

AC代码：

 1 #include<cstdio> 
 2 #define MAXM 2710
 3 #define MAXV 505
 4 #define inf 1<<29
 5 
 6 struct node{
 7     int x,y,t;
 8 }edge[MAXM];
 9 
10 int n,m,w;//n区域个数 m条路 w个虫洞 
11 
12 int bellman_ford()
13 {
14     int i,j,d[MAXV],flag=1,cnt=1;
15     for(i=1;i<=n;i++) 
16         d[i]=inf;//出发点到i的距离全部初始化为无穷大 
17     while(flag)
18     {
19         flag=0;//是为了把所有路走一遍 
20         if(cnt++>n) 
21             return 1;
22         //i是路的标号 
23         for(i=1;i<=m;i++)//普通道路（非虫洞）找最短路 
24         {
25             //分别找到edge[i].x和edge[i].y的最短距离 
26             if(d[edge[i].x]+edge[i].t < d[edge[i].y])  
27                 {d[edge[i].y] = d[edge[i].x]+edge[i].t;     flag=1;}
28             if(d[edge[i].y]+edge[i].t < d[edge[i].x]) 
29                 {d[edge[i].x] = d[edge[i].y]+edge[i].t;        flag=1;}
30         }
31         for(;i<=m+w;i++)//虫洞找最短路 
32             if(d[edge[i].y] > d[edge[i].x]-edge[i].t) 
33                 {d[edge[i].y] = d[edge[i].x]-edge[i].t;        flag=1;}
34     }
35     return 0;
36 }
37 
38 int main()
39 {
40     int t,i;
41     scanf("%d",&t);
42     while(t--)
43     {
44         scanf("%d%d%d",&n,&m,&w);
45         for(i=1;i<=m+w;i++)
46             scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].t);
47         if(bellman_ford()) 
48             printf("YES
");
49         else 
50             printf("NO
");
51     }
52     return 0;
53 }