Cow Relays(Floyed+倍增)

最新推荐文章于 2022-03-07 16:44:31 发布

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最新推荐文章于 2022-03-07 16:44:31 发布

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原文链接：http://www.cnblogs.com/zyf3855923/p/9597019.html

本文介绍了一种利用矩阵快速幂的方法解决特定图上的最短路径问题。该问题源自一场牛牛接力赛，需要找到连接起点和终点且恰好经过N条路径的最短路线。文章提供了一个C++实现示例，通过倍增思想优化了搜索过程。

题目描述

For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.
Each trail connects two different intersections (1 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the lengthi of each trail (1 ≤ lengthi ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.
To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.
Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.

输入

* Line 1: Four space-separated integers: N, T, S, and E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: lengthi , I1i , and I2i

输出

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

样例输入

样例输出

10
倍增的思想，通过矩阵快速幂进行优化。

#include <bits/stdc++.h>
#define maxn 105
using namespace std;
typedef long long ll;
int cnt=1;
struct Matrix{
    int a[maxn][maxn];
    Matrix operator * (const Matrix &x) const
    {
        Matrix c;
        memset(c.a,0x3f,sizeof(c.a));
        for (int k=1; k<=cnt; k++)
            for (int i=1; i<=cnt; i++)
                for (int j=1; j<=cnt; j++)
                    c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]);
        return c;
     }
}s,ans;
void ksm(int n)
{
    ans = s;
    n--;
    for (; n; n>>=1)
    {
        if (n&1) ans = ans*s;
        s = s*s;
    }
}
int vis[1005];
int main()
{
   int n,m,S,t;
   //freopen("in.txt","r",stdin);
   scanf("%d%d%d%d",&n,&m,&S,&t);
   memset(s.a,0x3f,sizeof(s.a));
   for(int i=1;i<=m;i++)
   {
       int x,u,v;
       scanf("%d%d%d",&x,&u,&v);
       if(!vis[u])
       {
           vis[u]=++cnt;
       }
       if(!vis[v])
       {
           vis[v]=++cnt;
       }
       s.a[vis[u]][vis[v]]=s.a[vis[v]][vis[u]]=x;
   }
   ksm(n);
   printf("%d\n",ans.a[vis[S]][vis[t]]);
   return 0;
}

转载于:https://www.cnblogs.com/zyf3855923/p/9597019.html