HDU 4911 Inversion

          http://acm.hdu.edu.cn/showproblem.php?pid=4911   归并排序求逆对数。

                Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 556    Accepted Submission(s): 239

Problem Description

bobo has a sequence a₁,a₂,…,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≤10⁵,0≤k≤10⁹). The second line contains n integers a₁,a₂,…,a_n (0≤a_i≤10⁹).

 

 

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：相邻的最多调换k次，使得逆对数最小，

思路，先求出整个序列的逆对数-k次就可，如果出现负数就输出为0.用归并排序求逆对数。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
__int64 a[100005],cnt,c[100005];
__int64 k;
void merg(int low, int mid, int high)
{
  int i=low, j=mid+1;
  cnt=0;
  while(i<=mid&&j<=high)
  {
    if(a[i]>a[j])
    {
      c[cnt++]=a[j++];
      k+=mid-i+1;
    }
    else
    {
      c[cnt++]=a[i++];
    }
  }
  while(i<=mid)
  {
    c[cnt++]=a[i++];
  }
  while(j<=high)
  {
    c[cnt++]=a[j++];
  }
  cnt=0;i=low;
  while(i<=high)
  {
    a[i++]=c[cnt++];
  }
}
void merger(int low, int high)
{
  int mid;
  if(low<high)
  {
    mid=(low+high)/2;
    merger(low,mid);
    merger(mid+1,high);
    merg(low,mid,high);
  }
}
int main()
{
  int i,n,m;
  while(~scanf("%d%d",&n,&m))
  {
           k=0;
    for(i=0; i<n; i++)
    {
      scanf("%I64d",&a[i]);
    }
       merger(0,n-1);
    if(k-m<=0)
       printf("0\n");
     else
    printf("%I64d\n",k-m);
  }
  return 0;
}