| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 32109 | Accepted: 11660 |
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
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. 一个是单向边，而且单向边为负值。
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 5 const int VM=520; 6 const int EM=5020; 7 const int INF=0x3f3f3f3f; 8 9 struct Edge{ 10 int u,v; 11 int c; 12 }edge[EM<<1]; 13 14 int n, m, w, cnt, dis[VM]; 15 16 17 void add(int u, int v, int c) 18 { 19 edge[cnt].u = u; 20 edge[cnt].v = v; 21 edge[cnt].c = c; 22 cnt++; 23 } 24 int Bellman() 25 { 26 int i, j, k, flag; 27 for(i=0; i<n; i++) 28 dis[i] = INF; 29 for(i=1; i<n; i++) 30 { 31 flag = 0; 32 for(j=0; j<cnt; j++) 33 { 34 if(dis[edge[j].v] > dis[edge[j].u] + edge[j].c) 35 { 36 dis[edge[j].v] = dis[edge[j].u] + edge[j].c; 37 flag = 1; 38 } 39 } 40 if(!flag) break; 41 } 42 for(k=0; k<cnt; k++) 43 if(dis[edge[k].v] > dis[edge[k].u] + edge[k].c) 44 return 1; 45 return 0; 46 47 } 48 int main() 49 { 50 int T, i; 51 scanf("%d", &T); 52 while(T--) 53 { 54 int x, y, z; 55 cnt = 0; 56 scanf("%d%d%d", &n, &m, &w); 57 for(i=0; i<m; i++) 58 { 59 scanf("%d%d%d", &x, &y, &z); 60 add(x, y, z);//加边 61 add(y, x, z); 62 } 63 for(i=0; i<w; i++) 64 { 65 scanf("%d%d%d", &x, &y, &z); 66 add(x, y, -z);//注意这里是-z 67 } 68 if(Bellman()) 69 printf("YES "); 70 else printf("NO "); 71 } 72 return 0; 73 }