POJ---3259 Wormholes[SPFA模版]
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 20567 | Accepted: 7314 |
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 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
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 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.
Source
John的农场里N块地,M条路连接两块地,W个虫洞;路是双向的,虫洞是一条单向路,会在你离开之前把你传送到目的地,
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来,看到了离开之前的自己。
简化下,就是看图中有没有负权环。
1 #include<iostream> 2 #include<queue> 3 using namespace std; 4 5 #define MAXN 510 6 #define intmax 0x3fffffff 7 8 int map[MAXN][MAXN]; 9 int cnt[MAXN]; 10 int vst[MAXN]; 11 int dis[MAXN]; 12 int f,n,m,w; 13 14 bool spfa(int st) 15 { 16 int i; 17 for(i=0;i<=n;i++) 18 { 19 vst[i]=0; 20 dis[i]=intmax; 21 cnt[i]=0; 22 } 23 int temp; 24 queue<int>Que; 25 Que.push(st); 26 dis[st]=0; 27 vst[st]=1; 28 cnt[st]=1; 29 while(!Que.empty()) 30 { 31 temp=Que.front(); 32 Que.pop(); 33 vst[temp]=0; 34 for(i=1;i<=n;i++) 35 { 36 if(dis[i]>dis[temp]+map[temp][i]) 37 { 38 dis[i]=dis[temp]+map[temp][i]; 39 if(!vst[i]) 40 { 41 cnt[i]++; 42 Que.push(i); 43 vst[i]=1; 44 if(cnt[i]>=n) 45 return false; 46 } 47 } 48 } 49 } 50 return true; 51 } 52 53 int main() 54 { 55 int i,j; 56 int s,e,t; 57 scanf("%d",&f); 58 while(f--) 59 { 60 scanf("%d%d%d",&n,&m,&w); 61 for(i=1;i<=n;i++) 62 for(j=1;j<=n;j++) 63 map[i][j]=intmax; 64 for(i=1;i<=m;i++) 65 { 66 scanf("%d%d%d",&s,&e,&t); 67 if(t<map[s][e]) 68 { 69 map[s][e]=t; 70 map[e][s]=t; 71 } 72 } 73 for(i=1;i<=w;i++) 74 { 75 scanf("%d%d%d",&s,&e,&t); 76 map[s][e]=-t; 77 } 78 if(spfa(1)) 79 printf("NO\n"); 80 else 81 printf("YES\n"); 82 } 83 return 0; 84 }
If you have any questions about this article, welcome to leave a message on the message board.
Brad(Bowen) Xu
E-Mail : maxxbw1992@gmail.com