POJ-3613 Cow Relays (矩阵乘法Floyd+倍增思想快速幂）

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我是靠谱客的博主悲凉丝袜，最近开发中收集的这篇文章主要介绍POJ-3613 Cow Relays (矩阵乘法Floyd+倍增思想快速幂），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Cow Relays

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8810		Accepted: 3474

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

Sample Output

Source

USACO 2007 November Gold

#include<cstdio>
#include<algorithm>
#include<cstring>
#define INF 0x3f3f3f3f
#define MAXN 210
using namespace std;
int Hash[1010];
int tn=0;
int n,t,s,e;
//本题的意思就是：经过k条边的最短路径
//本题的k就是n了，n头牛的数目
//倍增优化+快速幂
struct Matrix         //矩阵乘法floyd
{
    int a[MAXN][MAXN];
    Matrix operator *(const Matrix &x) const
    {
        Matrix c;
        memset(c.a,INF,sizeof(c.a));
        for (int k=1;k<=tn;k++)
            for (int i=1;i<=tn;++i)
                for (int j=1; j<=tn; ++j)                           //这个是单步Floyd,一次插入一个点
                    c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]);   //矩阵乘法Floyd与正常的区别：
                                                                    //1.没有对原数组进行更新，即原来的两点距离并没有改变
        return c;                                                   //由第一点的区别可以推出第二点的区别
                                                                    //2.一定是加一个点，添一条边
                                                                    //一般的Floyd可能加一个点，添多条边，因为他不断dp,记录了之前的最优
                                                                    //而最优可能不是两点的直接路径，有可能是加点的转达，这样就造成了
                                                                    //加一个点添多条边
     }
}str,ans;
void pow_mod()
{
    ans=str;          //插入n-1个点，求到的则是经过n条边的最短路径
    n--;              //n--是意思算str^(n-1)次方吧
    while(n!=0)
    {
        if((n&1)==1)
        ans=ans*str;
        str=str*str;
        n>>=1;
    }
}
int main()
{
    scanf("%d%d%d%d",&n,&t,&s,&e);
    memset(str.a,INF,sizeof(str.a));
    int i;
    for(i=1;i<=t;i++)
    {                                       //顶点的上限是10^3，但立方的话，也还是会超。
                                            //但发现边只有最多100条
        int x,y,w;                          //最多有200个点可以用，所以才想到hash压缩下
        scanf("%d%d%d",&w,&x,&y);
        if (!Hash[x]) Hash[x]=++tn;         //这也就是邻接矩阵的离散吧，离散不是可以用邻接表吗，因为Floyd不是邻接表的数据结构
        if (!Hash[y]) Hash[y]=++tn;
        str.a[Hash[x]][Hash[y]]=str.a[Hash[y]][Hash[x]]=w;
    }
    pow_mod();
    printf("%d",ans.a[Hash[s]][Hash[e]]);
    return 0;
}