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  • Cow Relays(Floyed+倍增)

    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

    2 6 6 4
    11 4 6
    4 4 8
    8 4 9
    6 6 8
    2 6 9
    3 8 9
    

    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
    ",ans.a[vis[S]][vis[t]]);
       return 0;
    }
    

      

     
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  • 原文地址:https://www.cnblogs.com/zyf3855923/p/9597019.html
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