poj Wormholes 最短路径bellman_ford

Posted 2020-08-14 唐唐123

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.

 

【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\n");
49         else 
50             printf("NO\n");
51     }
52     return 0;
53 }