【題目描述】

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 8

Sample Output

NO YES

【題目分析】
很正常的判斷負環，而且就算有重邊也并不影響。
【AC代碼】

#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<cmath> #include<climits> #include<queue> #include<vector> #include<set> #include<map> using namespace std;typedef long long ll; const int MAXN=505; const int MAXM=6005; struct node {int u,v,w; }edge[MAXM],E; int tot; int n,m,k; int dis[MAXN];inline void AddEdge(int u,int v,int w) {edge[tot].u=u; edge[tot].v=v; edge[tot].w=w;tot++; }bool Bellman() {memset(dis,0x3f,sizeof(dis));dis[1]=0;bool flag;int T=n;while(T--){flag=false;for(int i=0;i<tot;i++){E=edge[i];if(dis[E.v]>dis[E.u]+E.w){dis[E.v]=dis[E.u]+E.w;flag=true;}}if(!flag) break;}return T==-1; //T的值最后是-1，我還以為是0.。。。 }int main() {int T,u,v,w;scanf("%d",&T);while(T--){scanf("%d%d%d",&n,&m,&k);tot=0;for(int i=0;i<m;i++){scanf("%d%d%d",&u,&v,&w);AddEdge(u,v,w); AddEdge(v,u,w);}for(int i=0;i<k;i++){scanf("%d%d%d",&u,&v,&w);AddEdge(u,v,-w);}if(Bellman()){printf("YES\n");}else{printf("NO\n");}}return 0; }

總結

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