POJ Wormholes 3259(最短路)

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本文介绍了一种通过路径和虫洞实现时间旅行的算法，利用Bellman-Ford和SPFA算法来判断是否能回到起点的时间点之前。适用于解决有关负权边和负权回路的问题。

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 43742		Accepted: 16061

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 Tseconds.

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

上代码

Bellman_Ford

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>
#include <iostream>
using namespace std;
#define inf 0x3f3f3f3f
int top,n;
int dist[1000];
struct node
{
    int u,v,w;
}q[6000];
int bellman()
{
    int i,j;
    for(i=1;i<=n;i++)
    {
        dist[i]=inf;
    }
    dist[1]=0;
    for(i=1;i<n;i++)
    {
        for(j=0;j<top;j++)
        {
            int u=q[j].u;
            int v=q[j].v;
            if( dist[v]>dist[u]+q[j].w)
            {
                dist[v]=dist[u]+q[j].w;
            }
        }
    }
    for(i=0;i<top;i++)
    {
            int u=q[i].u;
            int v=q[i].v;
            if( dist[v]>dist[u]+q[i].w)
            {
                return 0;
            }

    }
    return 1;
}
int main()
{
    int m,i,j,l,k;
    int t;
    int a,b,c;
    cin>>t;
    while(t--)
    {
        top=0;
        cin>>n>>m>>l;
        while(m--)
        {
            cin>>a>>b>>c;
            q[top].u=a;
            q[top].v=b;
            q[top++].w=c;
            q[top].u=b;
            q[top].v=a;
            q[top++].w=c;
        }
        while(l--)
        {
            cin>>a>>b>>c;
            q[top].u=a;
            q[top].v=b;
            q[top++].w=-c;
        }
       if(!bellman())
       {
           printf("YES\n");
       }
       else
       {
           printf("NO\n");
       }
    }
    return 0;
}

SPFA

#include <queue>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define inf 1<<26
struct node
{
    int v,w;
    int next;
}q[6000];
int head[6000],dist[6000],vis[6000],c[6000];
int top,n;
void chu()
{
    int i;
    top=0;
    for(i=0;i<=n;i++)
    {
        head[i]=-1;
        vis[i]=0;
        c[i]=0;
    }
}
void add(int u,int v,int w)
{
    q[top].v=v;
    q[top].w=w;
    q[top].next=head[u];
    head[u]=top++;
}
void spfa()
{
    int i,j;
    for(i=1;i<=n;i++)
    {
        dist[i]=inf;
    }
    vis[1]=1;
    dist[1]=0;
    c[1]++;
    queue<int >Q;
    Q.push(1);
    while(!Q.empty())
    {
        int x=Q.front();
        vis[x]=0;
        Q.pop();
        for(i=head[x];i!=-1;i=q[i].next)
        {
            int v=q[i].v;
            int w=q[i].w;
            if(dist[v]>dist[x]+w)
            {
                dist[v]=dist[x]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    c[v]++;
                    if(c[v]>=n)
                    {
                        cout<<"YES\n";
                        return ;
                    }
                    Q.push(v);
                }
            }
        }
    }
    cout<<"NO\n";
}
int main()
{
    int t,m,i,j,l;
    cin>>t;
    int a,b,c;
    while(t--)
    {

        cin>>n>>m>>l;
        chu();
        while(m--)
        {
            cin>>a>>b>>c;
            add(a,b,c);
            add(b,a,c);
        }
        while(l--)
        {
            cin>>a>>b>>c;
            add(a,b,-c);
        }
        spfa();

    }



    return 0;
}