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.
/* 判断负权回路 加开一个数组 use[] */ #include <cstdio> #include <cstdlib> #include <cstring> struct node { int x,y,n; }e[10000]; int k[1000]; int d[1000]; bool v[1000]; int f[10000000]; int use[1000]; int tot; int F; int N,M,W; void add(int a,int b,int c) { e[++tot].x=b; e[tot].y=c; e[tot].n=k[a]; k[a]=tot; } bool SPFA(int start) { int head=0,tail=1; f[tail]=start; d[start]=0; v[start]=true; while (head<tail) { int x=f[++head]; v[x]=false; int t=k[x]; while (t) { if (d[e[t].x]>(d[x]+e[t].y)) { d[e[t].x]=d[x]+e[t].y; if (!v[e[t].x]) { f[++tail]=e[t].x; v[e[t].x]=true; use[e[t].x]++; if (use[e[t].x]>N) { return true; } } } t=e[t].n; } } return false; } int main() { scanf("%d\n",&F); for (int i=1;i<=F;i++) { tot=0; memset(v,0,sizeof(v)); memset(e,0,sizeof(e)); memset(k,0,sizeof(k)); memset(d,63,sizeof(d)); memset(use,0,sizeof(use)); scanf("%d %d %d\n",&N,&M,&W); for (int j=1;j<=M;j++) { int a,b,c; scanf("%d %d %d\n",&a,&b,&c); add(a,b,c); add(b,a,c); } for (int j=1;j<=W;j++) { int a,b,c; scanf("%d %d %d\n",&a,&b,&c); add(a,b,-c); } bool ans=false; for (int j=1;j<=N;j++) { if (d[j]>1000000) { if (SPFA(j)) { ans=true; break; } } } if (ans) printf("YES\n");else printf("NO\n"); } }