D - Inversion

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最新推荐文章于 2025-02-06 22:07:03 发布

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分类专栏： Divide and Conquer / Binary 文章标签：归并排序

本文链接：https://blog.youkuaiyun.com/qq_42217376/article/details/81634726

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本文介绍了一种名为D-Inversion的问题解决方法，该问题旨在通过有限次相邻元素交换，找到序列最小逆序数减去k的值。文章提供了一段使用归并排序策略来计算逆序数的C++代码示例，并附带详细注释以便理解。

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D - Inversion

bobo has a sequence a 1,a 2,…,a n. He is allowed to swap two adjacent numbers for no more than k times.

Find the minimum number of inversions after his swaps.

Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and a i>aj.

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,k (1≤n≤10 5,0≤k≤10 9). The second line contains n integers a 1,a 2,…,a n (0≤a i≤10 9).

Output

For each tests:

A single integer denotes the minimum number of inversions.

Sample Input

Sample Output

1
2

这个题的意思就是找逆序数-k是几。

归并讲解：

https://blog.youkuaiyun.com/qq_42217376/article/details/81224001

会求逆序数后就好做了

AC代码：

#include<bits/stdc++.h>
#include<algorithm>
using namespace std;
long long a[100000],b[100000];
long long times; 
void Merge(int s,int m,int e)
{
	int p=0;
	int p1=s,p2=m+1;
	while(p1<=m&&p2<=e)
	{
		if(a[p1]<=a[p2])
			b[p++]=a[p1++];
		else
		{
			b[p++]=a[p2++];
			times+=m-p1+1;
		} 
	}
		while(p1<=m)
			b[p++]=a[p1++];
		while(p2<=e)
			b[p++]=a[p2++];
		for(int i=0;i<e-s+1;i++)
			a[s+i]=b[i];	
}
void Merge_sort(int s,int e)
{
	if(s<e)
	{
		int m=s+(e-s)/2;
		Merge_sort(s,m);
		Merge_sort(m+1,e);
		Merge(s,m,e);
	}
}
int main()
{
	long long n,k;
	while(~scanf("%I64d%I64d",&n,&k))
	{
		for(int i=0;i<n;i++)
		scanf("%lld",&a[i]);
		times=0;
		Merge_sort(0,n-1);
		if(times-k>0)
			printf("%I64d\n",times-k);
		else
			printf("0\n");
	}
	return 0;
}