POJ 3268 Silver Cow Party

最新推荐文章于 2021-04-01 18:48:34 发布

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本文介绍了一种通过两次SPFA算法解决最优路径问题的方法，适用于有向图中寻找从多个起点到指定终点再返回起点的最短往返路径问题。具体实现上，通过边的正反向标识来优化计算过程。

Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow’s return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

【题目分析】
巧妙地想法就是把边反过来跑一边，这要n次SPFA变成了2次。我用的方法是用和一的异或值来判断是正向与反向。

【代码】

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
int h[1001],ne[200001],to[200001],dis[1001],w[200001],en,inq[1001],ans[1001];
inline void add(int a,int b,int c)
{ne[en]=h[a];to[en]=b;w[en]=c;h[a]=en++;}
inline void SPFA(int s,int tmp)
{
    memset(dis,0x3f,sizeof dis);
    queue<int>q;
    inq[s]=1;q.push(s);dis[s]=0;
    while (!q.empty())
    {
        int x=q.front();q.pop();inq[x]=0;
        for (int i=h[x];i>=0;i=ne[i])
        {
            if (i%2==tmp) continue;
            if (dis[to[i]]>dis[x]+w[i])
            {
                dis[to[i]]=dis[x]+w[i];
                if (!inq[to[i]])
                {
                    inq[to[i]]=1;
                    q.push(to[i]);
                }
            }
        }
    }
}
int main()
{
    memset(h,-1,sizeof h);
    int n,m,k;
    cin>>n>>m>>k;
    for (int i=1;i<=m;++i)
    {
        int a,b,c;
        cin>>a>>b>>c;
        add(a,b,c);
        add(b,a,c);
    }
    SPFA(k,0);
//  for (int i=1;i<=n;++i) cout<<dis[i]<<" "; cout<<endl;
    memcpy(ans,dis,sizeof dis);
    SPFA(k,1);
//  for (int i=1;i<=n;++i) cout<<ans[i]<<" "; cout<<endl;
    int anss=0;
    for (int i=1;i<=n;++i) anss=max(anss,dis[i]+ans[i]);
    cout<<anss<<endl;
}