FZU Problem 1919 K-way Merging sort(大数+记忆化搜索)
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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,…,Lk of nearly equal length. Some Li’s (namely, n reminder k of them, so possibly none) will have length , let these have the low indices: L1,L2,…,Ln%k Other Li’s will have length , and high indices are assigned: Ln%k+1,…,Lk-1,Lk. We intend to recursively sort the Li’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 ≤ 10 100) 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
42 2
3 2
100 7
1000 10
Sample Input
Case 1: 1Case 2: 3
Case 3: 1085
Case 4: 22005
题目大意:
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?
解题思路:
对于
n
个元素k-路分治,将元素
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);
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