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 }