求逆序数

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/A

 

题目：

  

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

 

 

 

题意：

    允许交换k对相邻的数，求序列的逆序数。 

分析：

    直接求出原序列中的逆序数，用逆序数减去交换的次数k即可。

   注意当最终值小于0时将其换成0。

#include<iostream>
using namespace std;
const int maxn=100005;
 long long a[maxn],t[maxn];
long long count;
void slove( 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;
   slove(A,x,m,T);
   slove(A,m,y,T);
   while(p<m||q<y)
       if(q>=y||(p<m&&A[p]<=A[q]))
           T[i++]=A[p++];
       else  {
           T[i++]=A[q++];
           count+=m-p;
       }
      for(i=x;i<y;i++)  A[i]=T[i];
 }
}
int main()
{
   int n;
    long long k;
  while(cin>>n>>k)
  { 
      count=0;
   for(int i=0;i<n;i++)
       cin>>a[i];
   slove(a,0,n,t);
   if(k<count)
   cout<<count-k<<endl;
   else cout<<'0'<<endl;
  }
   return 0;
}