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FZU 1919 -- K-way Merging sort（记忆化搜索）

2018-06-17 21:49:45来源：未知阅读 ()

题目链接

Problem Description

As we all known, merge sort is an O(nlogn) comparison-based sorting algorithm. The merge sort achieves its good runtime by a divide-and-conquer strategy, namely, that of halving the list being sorted: the front and back halves of the list are recursively sorted separately, then the two results are merged into the answer list. An implementation is shown as follows:

The procedure Merge(L1,L2:in List_type;L:out List_type) that we have in mind for sorting two lists is described as follows. Initialize pointers to the first item in each list L1,L2, and then

repeat

compare the two items pointed at;

move the smaller into L;

move the pointer which originally points at the smaller one to the next number;

until one of L1,L2 exhausts;

drain the remainder of the unexhausted list into L;

Now let us come to the situation when there are k pointers, here k≥2. Let L be a list of n elements. Divide L into k disjoint contiguous sublists L₁,L₂,…,L_k of nearly equal length. Some L_i’s (namely, n reminder k of them, so possibly none) will have length , let these have the low indices: L₁,L₂,…,L_n%k Other L_i’s will have length , and high indices are assigned: L_n%k+1,…,L_k-1,L_k. We intend to recursively sort the L_i’s and merge the k results into an answer list.

We use Linear-Search-Merge here to merge k sorted lists. We find the smallest of k items (one from each of the k sorted source lists), at a cost of k-1 comparisons. Move the smallest into the answer list and advances its corresponding pointer (the next smallest element) in the source list from which it came. Again there are k items, from among which the smallest is to be selected. (When i (1 ≤ i < k) lists are empty, k-way merging sort becomes to (k-i)-way merging sort, and the draining process will start when the total order of all the elements have been found)

Given a list containing n elements, your task is to find out the maximum number of comparisons in k-way merging sort.

Input

The first line of the input contains an integer T (T <= 100), indicating the number of cases. Each case begins with a line containing two integer n (1 ≤ n ≤ 10¹⁰⁰) and k (2 ≤ k ≤ 20), the number of elements in the list, and it is k-way merging sort.

Output

For each test case, print a line containing the test case number (beginning with 1) and the maximum number of comparisons in k-way merging sort.

Sample Input

2 2

3 2

100 7

1000 10

Sample Output

Case 1: 1

Case 2: 3

Case 3: 1085

Case 4: 22005

题意：对于归并排序，本来是2-路分治，现在要求k-路分治，求n个元素下k-路分治归并排序所需要的最大比较次数？

思路：对于n个元素k-路分治，将元素1~n分为k段，前s=n%k段每段有n/k+1个元素，后k-s段每段有n/k个元素，那么对这k段进行归并时和2-路一样，选出这k段中的最小值放入合并后的数组，每次从k段中选出一个最小值需要比较k-1次，当每一段均剩下一个元素时，共选出了n-k个最小值，所以比较了（k-1）*（n-k）次，对于剩下的k个元素则需要比较（k-1）+（k-2）+……+1=（k-1）*k/2 次，所以共需要（k-1）*（n-k）+（k-1）*k/2次，然后递归进一步分治，注意使用记忆化搜索。

注意：本题数据很大所以不能用数组存值，而用的map映射的。

代码如下：

import java.math.BigInteger;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main{
    static Map<BigInteger,BigInteger>dp = new HashMap<BigInteger,BigInteger>();
    static BigInteger n, ans;
    static int k;
    static BigInteger dfs(BigInteger len, BigInteger x){
        if(dp.containsKey(len)) return x.multiply(dp.get(len));
        if(len.compareTo(BigInteger.valueOf(k))<=0){
            return x.multiply(len.subtract(BigInteger.ONE)).multiply(len).divide(BigInteger.valueOf(2));
        }
        BigInteger tmp = (BigInteger.valueOf(k).subtract(BigInteger.ONE)).multiply((len.subtract(BigInteger.valueOf(k))));
        tmp = tmp.add(BigInteger.valueOf(k).multiply(BigInteger.valueOf(k).subtract(BigInteger.ONE)).divide(BigInteger.valueOf(2)));
        BigInteger kk = len.mod(BigInteger.valueOf(k));
        if(kk!=BigInteger.ZERO){
            tmp=tmp.add(dfs(len.divide(BigInteger.valueOf(k)).add(BigInteger.ONE),kk));
        }
        tmp = tmp.add(dfs(len.divide(BigInteger.valueOf(k)),BigInteger.valueOf(k).subtract(kk)));
        dp.put(len, tmp);
        return tmp.multiply(x);
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for(int cas=1; cas<=T; cas++){
            dp.clear();
            n = in.nextBigInteger();
            k = in.nextInt();
            ans=dfs(n, BigInteger.ONE);
            System.out.println("Case "+cas+": "+ans);
        }
    }
}

下面这个是我先用c++写的版本，用来验证算法，上面的java是用这c++代码改写的（其实一样）。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <map>
using namespace std;
typedef long long LL;
LL n,k;
map<LL,LL>mp;

LL dfs(LL len,LL x)
{
    if(mp.count(len)) return x*mp[len];
    if(len<=k) return x*(len-1)*len/2;
    LL tmp=(k-1)*(len-k);
    tmp+=k*(k-1)/2;
    tmp+=dfs(len/k+(len%k!=0),(len%k));
    tmp+=dfs(len/k,(k-len%k));
    mp[len]=tmp;
    return tmp*x;
}

int main()
{
    while(scanf("%lld%lld",&n,&k)!=EOF)
    {
        mp.clear();
        LL ans=dfs(n,1);
        cout<<"final ans = "<<ans<<endl;
    }
    return 0;
}