POJ 3259——Wormholes 图论

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#算法 #poj #图论 #最短路径

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本文探讨了农民FJ利用农场中的时空隧道和双向路径，尝试在其出发时间之前返回起点的问题。通过应用Bellman-Ford算法检测负环来判断是否能够实现时间旅行的目标，为读者揭示了图论在解决复杂问题中的巧妙应用。

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原题如下

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

Sample Output

NO
YES

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.

题目大意：一个农夫，在他的农场里有m条双向路径和w个单向虫洞。经过每条路需要耗费t秒时间，穿越每个虫洞可以使时间倒退t秒。求他能否在他出发时间之前到达出发地点？

分析：显然是图论相关问题。一个虫洞就相当于负的边权。要想达到要求，必须得通过某个顶点两次。但是如果不存在负圈，那么最短路最多经过n-1条边，在bellman-ford算法中，就相当于只能更新n-1次，也不可能经过同一个顶点两次。

因此题目也就转化成了，图中是否存在负圈。

代码如下：

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <queue>
#include <ctype.h>
#include <vector>
using namespace std;
const int MAXN=10000;
const int INF=1000000;

struct edge{
    int from,to,cost;
};
edge es[MAXN];
int d[MAXN];
int f,n,m,w;

bool find_negative_loop(){
    memset(d,0,sizeof(d));
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<2*m+w;j++)
        {
            edge e=es[j];
            if(d[e.to]>d[e.from]+e.cost)
            {
                d[e.to]=d[e.from]+e.cost;
                if(i==n-1)
                    return true;
            }
        }
    }
    return false;
}
int main()
{
    cin>>f;
    while(f--)
    {
        cin>>n>>m>>w;
    int s,e,t;
    for(int i=0;i<2*m;)
    {
        cin>>s>>e>>t;
        es[i].from=s;
        es[i].to=e;
        es[i].cost=t;
        i++;
        es[i].from=e;
        es[i].to=s;
        es[i].cost=t;
        i++;
    }
    for(int i=2*m;i<w+2*m;i++)
    {
        cin>>s>>e>>t;
        es[i].from=s;
        es[i].to=e;
        es[i].cost=-t;
    }
    if(find_negative_loop())
        printf("YES\n");
    else
        printf("NO\n");
    }
}