Wormholes 分类： POJ 2015-07-14 20:21 21人阅读 评论(0) 收藏

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35235		Accepted: 12861

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

有一些道路和虫洞，经过道路需要一定的时间，穿过虫洞可以会回到之前的时间，问是不是可以回到进入之前的时间

就是判断有没有负环SPFA+前向星 算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
#define exp 1e-9
#define INF 0x3f3f3f3f

using namespace std;
const int Max=550;
struct node
{
    int v;
    int w;
    int next;

} Map[Max],Head[10000];
int Dis[Max];
bool vis[Max];
int path[Max];
int Du[Max];
int n,m,w;
int top;
bool SPFA(int star)
{
    queue<int >Q;
    memset(Dis,INF,sizeof(Dis));
    memset(vis,false,sizeof(vis));
    memset(Du,0,sizeof(Du));
    memset(path,1,sizeof(path));
    Q.push(star);
    vis[star]=true;
    Du[star]++;
    Dis[star]=0;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        if(Du[u]>n)
            return true;
        vis[u]=false;
        int p=Map[u].next;
        while(p!=-1)
        {
            if(Dis[Head[p].v]>Dis[u]+Head[p].w)
            {
                Dis[Head[p].v]=Dis[u]+Head[p].w;
                if(!vis[Head[p].v])
                {
                    Q.push(Head[p].v);
                    Du[Head[p].v]++;
                    vis[Head[p].v]=true;
                }
            }
            p=Head[p].next;
        }
    }
    return false;
}
void Creat(int u,int v,int ww)
{
    Head[top].next=Map[u].next;
    Map[u].next=top;
    Head[top].v=v;
    Head[top].w=ww;
    top++;
}

int main()
{
    int T;
    int u,v,ww;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d",&n,&m,&w);
        top=0;
        for(int i=0; i<=n; i++)
        {
            Map[i].next=-1;
        }
        for(int i=0; i<m; i++)
        {
            scanf("%d %d %d",&u,&v,&ww);
            Creat(u,v,ww);
            Creat(v,u,ww);
        }
        for(int i=0; i<w; i++)
        {
            scanf("%d %d %d",&u,&v,&ww);
            Creat(u,v,-ww);

        }
        if(SPFA(1))
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
    return 0;
}

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