Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 31425 | Accepted: 11431 |
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 题目意思：john有n块农场（用1--n表示），农场之间有m条路径，每条路径有个权值为经过这条路径花费的时间，路径是双向的，农场中有w个虫洞（看过时间简史就知道了），虫洞表示为a,b,t即为从a农场到b农场花费 -t 时间（虫洞是单向的）。问john从某个农场出发再次回到该农场时时间是否是出发之前（有点绕嘴。。就是john从一个农场出发绕了一大圈再次回到农场时，此时的时间为出发之前即回到了过去）。 思路：判断负环即可。 代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 using namespace std;
8
9 #define N 505
10 #define inf 999999999
11
12 struct node{
13 int y, t;
14 };
15
16 int dis[N];
17 int visited[N];
18 int num[N];
19 vector<node>ve[N];
20 int n, m, w;
21
22 void init(){
23 int i;
24 for(i=0;i<=n;i++){
25 ve[i].clear();
26 dis[i]=inf;
27 }
28 memset(visited,0,sizeof(visited));
29 memset(num,0,sizeof(num));
30 }
31
32 int SPFA(int st){
33
34 int i, j, u;
35 node p, q;
36 queue<int>Q;
37 Q.push(st);
38 dis[st]=0;
39 visited[st]=1;
40 while(!Q.empty()){
41 u=Q.front();
42 Q.pop();
43 visited[u]=0;
44 for(i=0;i<ve[u].size();i++){
45 p=ve[u][i];
46 if(dis[p.y]>dis[u]+p.t){
47 dis[p.y]=dis[u]+p.t;
48 if(!visited[p.y]){
49 visited[p.y]=1;
50 Q.push(p.y);
51 }
52 num[p.y]++;
53 if(num[p.y]>=n) return 1;
54 }
55 }
56 }
57 return 0;
58 }
59
60 main()
61 {
62 int i, j, k, t;
63 int x, y, z;
64 cin>>t;
65 while(t--){
66 scanf("%d %d %d",&n,&m,&w);
67 init();
68 node p;
69 for(i=0;i<m;i++){
70 scanf("%d %d %d",&x,&y,&z);
71 p.y=y;p.t=z;
72 ve[x].push_back(p);
73 p.y=x;;
74 ve[y].push_back(p);
75 }
76 for(i=0;i<w;i++){
77 scanf("%d %d %d",&x,&y,&z);
78 p.y=y;p.t=-z;
79 ve[x].push_back(p);
80 }
81 if(SPFA(1))
82 printf("YES\n");
83 else printf("NO\n");
84 }
85 }