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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 34934 | Accepted: 12752 |
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
Hint
1 #include <iostream> 2 using namespace std; 3 struct farm { 4 int S; 5 int E; 6 int T; 7 } f[5500]; 8 int main() { 9 int num; 10 int N, M, W; 11 cin >> num; 12 int F[500]; 13 for (int i = 0; i < num; i++) { 14 cin >> N >> M >> W; 15 for (int j = 0; j < N; j++) { 16 F[j] = 20000; 17 } 18 F[0] = 0; 19 for (int j = 0; j < M; j++) { 20 int a, b, c; 21 cin >> a >> b >> c; 22 f[2*j].S = a; 23 f[2*j].E = b; 24 f[2*j].T = c; 25 f[2*j+1].S = b; 26 f[2*j+1].E = a; 27 f[2*j+1].T = c; 28 29 } 30 for (int j =2* M; j < 2*M + W; j++) { 31 int a, b, c; 32 cin >> a >> b >> c; 33 f[j].S = a; 34 f[j].E = b; 35 f[j].T = 0 - c; 36 } 37 for (int j = 0; j < N-1; j++) { 38 for (int k = 0; k < 2*M + W; k++) { 39 if (F[f[k].E] > F[f[k].S] + f[k].T) { 40 F[f[k].E] = F[f[k].S] + f[k].T; 41 } 42 } 43 } 44 int flag = 0; 45 for (int k = 0; k < 2*M + W; k++) { 46 if (F[f[k].E] >F[f[k].S] + f[k].T) { 47 F[f[k].E] = F[f[k].S] + f[k].T; 48 flag=1; 49 break; 50 } 51 } 52 if(flag){ 53 cout<<"YES"<<endl; 54 }else{ 55 cout<<"NO"<<endl; 56 } 57 } 58 return 0; 59 }
Wormholes - poj 3259 (Bellman-Ford算法)
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原文地址：http://www.cnblogs.com/sdxk/p/4626178.html
