POJ 3259 Wormholes 贝尔曼福特算法判负环
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Memory Limit: 65536KTotal Submissions: 32449
Accepted: 11791
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
23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8
Sample Output
NOYES
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
题目大意:John有几个农场,农场之间有路,也有虫洞,虫洞可以让时光倒流,John从某一点出发,看看是否能回到原来的位置并且能看到出发时的自己。意思就是让时光倒流,能回到他出发前。。。。判负环。。。
又是一个判负环的题。。。。。
#include <stdio.h>#include <string.h>struct node{ int u; int v; int w;} ls[10010];int num;const int inf = 99999999;void creat(int s,int e,int t){ ls[num].u = s; ls[num].v = e; ls[num].w = t; num++;}bool bf(int n){ bool flag; int dis[710]; for(int i = 0; i < n; i++) dis[i] = inf; for(int i = 1; i <= n - 1; i++) { flag = false; for(int j = 0; j < num; j++) { int u = ls[j].u; int v = ls[j].v; int w = ls[j].w; if(dis[v] > dis[u] + w) { flag = true; dis[v] = dis[u] + w; } } if(!flag) break; } flag = false; for(int i = 0; i < num; i++) { int u = ls[i].u; int v = ls[i].v; int w = ls[i].w; if(dis[v] > dis[u] + w) { flag = true; break; } } return flag;}int main(){ int n,m,k,s,e,t,w; while(~scanf("%d",&k)) { while(k--) { num = 0; scanf("%d%d%d",&n,&m,&w); for(int i = 0; i < m; i++) { scanf("%d%d%d",&s,&e,&t); creat(s,e,t); creat(e,s,t); } for(int i = 0; i < w; i++) { scanf("%d%d%d",&s,&e,&t); creat(s,e,-t); } if(bf(n)) printf("YES\n"); else printf("NO\n"); } } return 0;}
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