POJ 3259 Wormholes(SPFA算法判断是否存在负环)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 36755 Accepted: 13457

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, andW (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 toF (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 fromS toE that also moves the traveler backT 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

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 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：有n个地点，编号1到n。  有m行S ，E， T。表示从地点S(E)到地点E(S)的时间为T。  有w个虫洞，通过虫洞可以从S穿越到到E（单向穿越），且时间回到T秒前。问是否存在从一点出发，回到这一点看到从前的自己。

 

解题思路：先建图，能从虫洞穿越的路径的权值可以看成负值，若要回到起始点，且时间倒退，路径一定是至少经过一条负权值边的环，解答此题即为判断是否存在负值环即可。

 

SPFA算法，代码如下：

 

 

#include<cstdio>#include<queue>using namespace std;#define INF 0x3f3f3f#define maxn 510#define maxm 5510int dis[maxn],visit[maxn],head[maxn],top,n;struct node{int to,val,next;}edge[maxm];void add(int a,int b,int c){edge[top].to=b;edge[top].val=c;edge[top].next=head[a];head[a]=top++;}void spfa(){int mark[maxn];//记录一个点入队列的次数 int i,u,v;for(i=1;i<=n;++i){mark[i]=0;dis[i]=INF;visit[i]=0;}queue<int>q;q.push(1);dis[1]=0;visit[1]=1;mark[1]++;while(!q.empty()){u=q.front();q.pop();visit[u]=0;for(i=head[u];i!=-1;i=edge[i].next){v=edge[i].to;if(dis[v]>dis[u]+edge[i].val){dis[v]=dis[u]+edge[i].val;if(!visit[v]){visit[v]=1;mark[v]++;q.push(v);if(mark[v]>=n)//图中有负环 {printf("YES\n");return ;}}}}}printf("NO\n");}int main(){int m,w,a,b,c,i,t;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&w);top=0;for(i=1;i<=n;++i)   head[i]=-1;while(m--){scanf("%d%d%d",&a,&b,&c);add(a,b,c);add(b,a,c);}while(w--){scanf("%d%d%d",&a,&b,&c);add(a,b,-c);}spfa();}return 0;}