题目链接：http://poj.org/problem?id=3259
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 55082 | Accepted: 20543 |
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 #define rep(i,a,n) for(int (i) = a; (i)<=(n); (i)++) 13 #define ms(a,b) memset((a),(b),sizeof((a))) 14 using namespace std; 15 typedef long long LL; 16 const double EPS = 1e-8; 17 const int INF = 2e9; 18 const LL LNF = 9e18; 19 const int MOD = 1e9+7; 20 const int MAXN = 1e3+10; 21 22 int N, M, W; 23 24 struct edge 25 { 26 int to, w, next; 27 }edge[MAXN*MAXN]; 28 int cnt, head[MAXN]; 29 30 void add(int u, int v, int w) 31 { 32 edge[cnt].to = v; 33 edge[cnt].w = w; 34 edge[cnt].next = head[u]; 35 head[u] = cnt++; 36 } 37 38 void init() 39 { 40 cnt = 0; 41 memset(head, -1, sizeof(head)); 42 } 43 44 int dis[MAXN], times[MAXN], inq[MAXN]; 45 bool spfa(int st) 46 { 47 memset(inq, 0, sizeof(inq)); 48 memset(times, 0, sizeof(times)); 49 for(int i = 1; i<=N; i++) 50 dis[i] = INF; 51 52 queue<int>Q; 53 Q.push(st); 54 inq[st] = 1; 55 dis[st] = 0; 56 while(!Q.empty()) 57 { 58 int u = Q.front(); 59 Q.pop(); inq[u] = 0; 60 for(int i = head[u]; i!=-1; i = edge[i].next) 61 { 62 int v = edge[i].to; 63 if(dis[v]>dis[u]+edge[i].w) 64 { 65 dis[v] = dis[u]+edge[i].w; 66 if(!inq[v]) 67 { 68 Q.push(v); 69 inq[v] = 1; 70 if(++times[v]>N) return true; 71 } 72 } 73 } 74 } 75 return false; 76 } 77 78 int main() 79 { 80 int T; 81 scanf("%d",&T); 82 while(T--) 83 { 84 init(); 85 scanf("%d%d%d", &N, &M, &W); 86 for(int i = 1; i<=M; i++) 87 { 88 int s, e, t; 89 scanf("%d%d%d", &s, &e, &t); 90 add(s, e, t); 91 add(e, s, t); 92 } 93 94 for(int i = 1; i<=W; i++) 95 { 96 int s, e, t; 97 scanf("%d%d%d", &s, &e, &t); 98 add(s, e, -t); 99 } 100 101 int flag = 0; 102 for(int i = 1; i<=N; i++) 103 if(spfa(i)) 104 { 105 flag = 1; 106 break; 107 } 108 109 110 if(flag) 111 puts("YES"); 112 else 113 puts("NO"); 114 } 115 }