标签：des style blog color io os ar strong for
| 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
Output
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 }
标签：des style blog color io os ar strong for
原文地址：http://www.cnblogs.com/qq1012662902/p/3965425.html
