FZU Problem 1919 K-way Merging sort（大數+記憶化搜索）

傳送門

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

4
2 2
3 2
100 7
1000 10

Sample Input

Case 1: 1
Case 2: 3
Case 3: 1085
Case 4: 22005

題目大意：
對於歸併排序，本來是2-路分治，現在要求k-路分治，求n個元素下k-路分治歸併排序所需要的最大比較次數？

解題思路：
對於 n 個元素k-路分治，將元素 1−n 分爲 k 段，前s=n%k 段每段有nk+1 個元素，後 k−s 段每段有nk 個元素，那麼對這 k 段進行歸併時和2-路一樣，選出這 k 段中的最小值放入合併後的數組，每次從k段中選出一個最小值需要比較 k−1 次，當每一段均剩下一個元素時，共選出了n-k個最小值，所以比較了（k−1）∗（n−k） 次，對於剩下的 k 個元素則需要比較（k−1）+（k−2）+……+1=（k−1）∗k2 次，所以共需要（k−1）∗（n−k）+（k−1）∗k2 次，然後遞歸進一步分治，注意使用記憶化搜索，否則超時。

注意：本題數據很大所以不能用數組存值，而用的 map 映射的。

代碼如下：

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);
        }
    }
}