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: A single integer, F. F farm descriptions follow. 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
Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).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 8Sample Output
NO YESHint
For farm 1, FJ cannot travel back in time. 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. 题意：给出n个点和m条双向边且权值为正，w条单向边，权值为负。要求给定一个图，判断图中是否有负环。 题意：判断是否有环显然需要使用Bellman-Ford算法。因为本题只需判断是否有负环的存在，而不需求最短路，所以可令初始dis均为0，如果第n次松弛成功，则有负环；否则没有。 #include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> using namespace std; #define inf 0x3f3f3f3f const int maxn=550; double dis[maxn]; int tot;//边的总数 int way[maxn*maxn][2];//存边的起点和终点 int value[maxn*maxn];//存所花费的时间 int n,m,w; bool Bellman() { for(int i=1;i<=n;i++) dis[i]=0; for(int i=0;i<=n;i++)//循环n-1次 { bool flag=false;//优化 for(int j=0;j<tot;j++) { if(dis[way[j][1]]>dis[way[j][0]]+value[j]) { flag=true; dis[way[j][1]]=dis[way[j][0]]+value[j]; } } if(!flag)return false;//无环 } return true; } int main() { int t; scanf("%d",&t); int a,b,c; while(t--) { scanf("%d%d%d",&n,&m,&w); tot=0; for(int i=0;i<m;i++) { scanf("%d%d%d",&a,&b,&c); way[tot][0]=a; way[tot][1]=b; value[tot]=c; tot++; way[tot][0]=b; way[tot][1]=a; value[tot]=c; tot++; } for(int i=0;i<w;i++) { scanf("%d%d%d",&a,&b,&c); way[tot][0]=a; way[tot][1]=b; value[tot]=-c; tot++; } int i; if(Bellman()) printf("YES\n"); else printf("NO\n"); } return 0; }