ACM求经过k次移动后所得的逆序数

Description

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>a _j.

 

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

3 1

2 2 1

3 0

2 2 1

 

Sample Output

1

2

 

解题思路：

题目大意是输入一组数据通过k次相邻位置的移动后，求 他所得的最小的逆序数。我们通过归并的方法求出这一组数据的逆序数，然后将这个逆序数减去所要移动的次数就得到了我们要求的最小的逆序数。

程序代码：

#include <iostream>
using namespace std;
int n;
long long s;
long long a[100005],t[100005];
void fun(long long *a,int x,int y,long long*t)
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;
        int p=x,q=m,i=x;
        fun(a,x,m,t);
        fun(a,m,y,t);
        while(p<m||q<y)
        {
            if(q>=y||(a[p]<=a[q]&&p<m))
                t[i++]=a[p++];
            else
            {
                t[i++]=a[q++];
                s=s+m-p;
            }
        }
        for( i=x;i<y;i++)
            a[i]=t[i];
    }
}
int main()
{
    long long k;
    while(cin>>n>>k)
    {
        s=0;
        for(int i=0;i<n;i++)
            cin>>a[i];
        fun(a,0,n,t);
        if(s-k<=0)
            cout<<"0"<<endl;
        else
            cout<<s-k<<endl;
    }
    return 0;
}