poj 3259 Wormholes

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本文介绍了一种通过Bellman-Ford算法判断是否存在负权回路的方法，以解决一个有趣的问题：农夫约翰是否能利用农场中的路径和虫洞实现回到过去的时间旅行。

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题目链接：点击打开链接

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 requiresT 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

Sample Output

NO
YES

题目大意：有几个牧场，每个牧场有n个地方，m条路径，路径有权值是走的秒数．w个虫洞，虫洞可以穿越回前几秒，给出n个点m条正边　w条负边．

基本思路：判断有无负权回路，用bellman-ford判断，看一下在执行完算法后还能不能在松弛，如果能就表示由回路

<span style="font-size:18px;">#include<iostream>
#include<cstring>
#define INF 0x3f3f3f3f

using namespace std;

struct node
{
    int a,b,c;
} q[5500];

int dist[5500];
int top;

bool bellman(int n)
{
    for(int i=1; i<=n; i++)
    {
        dist[i]=INF;
    }
    dist[1]=0;
    for(int i=1; i<n; i++)
    {
        bool flag=false;
        for(int j=0; j<top; j++)
        {
            int v=q[j].a;
            int u=q[j].b;
            if(dist[v]>dist[u]+q[j].c)
            {
                dist[v]=dist[u]+q[j].c;
                flag=true;
            }
        }
        if(!flag) return false;
    }

    for(int j=0; j<top; j++)
    {
        if(dist[q[j].a]>dist[q[j].b]+q[j].c)
            return true;
    }
    return false;
}
int main()
{
    int n,m,w;
    int s,e,t;
    int k;
    cin>>k;
    while(k--)
    {
        memset(q,0,sizeof(q));
        cin>>n>>m>>w;
        top=0;
        for(int i=1; i<=m; i++)
        {
            cin>>s>>e>>t;
            q[top].a=s;
            q[top].b=e;
            q[top++].c=t;

            q[top].a=e;
            q[top].b=s;
            q[top++].c=t;
        }
        for(int i=1; i<=w; i++)
        {
            cin>>s>>e>>t;
            q[top].a=s;
            q[top].b=e;
            q[top++].c=-t;
        }
        if(bellman(n))cout<<"YES\n";
        else cout<<"NO\n";
    }
    return 0;
}

</span>