Wormholes
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.
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.
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 8Sample Output
NO YESHint
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.
John有n个农场,m条路径,w个虫洞。现在他想进行一个奇妙的旅行,使他能通过虫洞回到以前。问他能否实现。
一道Ballman—Ford模板题,判负环。
代码:
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; #define inf 0x3f3f3f #define V 505 #define N 5205 struct node { int u, v, t; }e[N]; int dis[V]; int k,n; int B_Ford(int s) { for(int i=1;i<=n;i++) //初始化 dis[i]=inf; dis[s]=0; //起点为0 for(int i=1;i<n;i++) { int flag=1; for(int j=0;j<k;j++) if(dis[e[j].v]>dis[e[j].u]+e[j].t) { dis[e[j].v]=dis[e[j].u]+e[j].t; flag=0; } if(flag) return 0; //如果不能松弛了,直接判断没有负环 } for(int i=0;i<k;i++) if(dis[e[i].v]>dis[e[i].u]+e[i].t) return 1; //如果还能松弛则存在负环 return 0; } int main() { int T, m, t, u, v, time; scanf("%d", &T); while(T--) { scanf("%d%d%d", &n, &m, &t); k = 0; for(int i=0; i<m; i++) { scanf("%d%d%d", &u, &v, &time); //路径为双向 e[k].u = u; e[k].v = v; e[k++].t = time; e[k].u = v; e[k].v = u; e[k++].t = time; } for(int i=0; i<t; i++) { scanf("%d%d%d", &u, &v, &time); //虫洞为单向路径 e[k].u = u; e[k].v = v; e[k++].t = -time; } if(B_Ford(1)) printf("YES\n"); else printf("NO\n"); } return 0; }