【题解】poj3613[USACO 2007 November Gold].Cow Relays floyd+矩阵快速幂

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最新推荐文章于 2022-12-28 19:31:32 发布

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分类专栏： USACO 算法竞赛进阶指南 poj 最短路问题矩阵乘法快速幂文章标签：矩阵快速幂 floyd

本文链接：https://blog.youkuaiyun.com/qq_41958841/article/details/82658460

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本文介绍了一道有趣的算法题——牛牛接力赛，通过使用矩阵快速幂结合Floyd算法来解决寻找从起点到终点恰好经过N条路径的最短距离问题。该题需要对图论中的最短路径算法有深刻理解。

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题目链接

Description

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.

Input

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

Output

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

Sample Input

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

Sample Output

矩阵A^m保存任意两点之间恰好经过m条边的最短路，采用矩阵快速幂+floyd

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
int n,pos,t,s,e;
map<int,int>mp;
struct MM{
    int ma[210][210];
}d,ans;
MM mul(MM ta,MM tb)
{
    MM c;
    memset(c.ma,0x3f,sizeof(c.ma));
    for(int i=1;i<=pos;i++)
        for(int j=1;j<=pos;j++)
            for(int k=1;k<=pos;k++)
                c.ma[i][j]=min(c.ma[i][j],ta.ma[i][k]+tb.ma[k][j]);
    return c; 
}
void qpow()
{
    ans=d;
    for(n=n-1;n;n>>=1)
    {
        if(n&1)ans=mul(ans,d);
        d=mul(d,d);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d%d%d%d",&n,&t,&s,&e);
    memset(d.ma,0x3f,sizeof(d.ma));
    int u,v,w;
    for(int i=1;i<=t;i++)
    {
        scanf("%d%d%d",&w,&u,&v);
        if(mp[u])u=mp[u];else u=mp[u]=++pos;
        if(mp[v])v=mp[v];else v=mp[v]=++pos;
        d.ma[u][v]=d.ma[v][u]=w;
    }
    qpow();
    printf("%d\n",ans.ma[mp[s]][mp[e]]);
    return 0;
}

总结

想到转化为矩阵非常关键