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  • poj 3259-- Wormholes(SPFA)

                                                                                                               Wormholes
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 37415   Accepted: 13764

    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 8

    Sample Output

    NO
    YES
    
    
    
    #include<stdio.h>
    #include<string.h> 
    #include<queue>
    #include<algorithm>
    using namespace std;
    #define INF 0x3f3f3f
    int m,n,cnt,p;
    int head[1010],vis[1010],used[1010],dist[1010];
    struct node
    {
    	int u,v;
    	int val,next;
    }edge[20010];
    void add(int u,int v,int val)
    {
    	edge[cnt].u=u;
    	edge[cnt].v=v;
    	edge[cnt].val=val;
    	edge[cnt].next=head[u];
    	head[u]=cnt++;
    }
    int SPFA(int st)
    {
    	queue<int>q;
    	memset(dist,INF,sizeof(dist));
    	memset(used,0,sizeof(used));
    	memset(vis,0,sizeof(vis));
    	vis[st]=1;
    	dist[st]=0;
    	q.push(st);//各种初始化 
    	while(!q.empty())
    	{
    		int u=q.front();
    		q.pop();
    		vis[u]=0;//进入循环一定要改成0,这样才能判断进入了多少次 
    		for(int i=head[u];i!=-1;i=edge[i].next)
    		{
    			int v=edge[i].v;
    			if(dist[v]>dist[u]+edge[i].val)
    			{
    				dist[v]=dist[u]+edge[i].val;
    				if(!vis[v])
    				{
    					vis[v]=1;
    					used[v]++;
    					if(used[v]>=m)//判断是否出现了负权边 
    					return 1;
    					q.push(v);
    				}
    			}
    		}
    	}
    	return 0;
    }
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		memset(head,-1,sizeof(head));
    		scanf("%d%d%d",&m,&n,&p);
    		int x,y,z;
    		while(n--)
    		{
    			scanf("%d%d%d",&x,&y,&z);
    			add(x,y,z);
    			add(y,x,z);
    		}
    		while(p--)
    		{
    			scanf("%d%d%d",&x,&y,&z);
    			add(x,y,-z);
    			add(y,x,INF);//虫洞是单向的,所以反向是无穷 
    		}
    		if(SPFA(1)) printf("YES
    ");
    		else printf("NO
    ");
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273806.html
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