Inversion_归并排序

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4911

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

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=100005;
__int64 cnt,k;
int a[N],c[N];
//将有二个有序数列a[first...mid]和a[mid...last]合并。
void merge(int a[],int first,int mid,int last,int c[])
{
    int i=first,j=mid+1;
    int m=mid,n=last;
    int k=0;
    while(i<=m||j<=n)
    {
        if(j>n||(i<=m&&a[i]<=a[j]))
            c[k++]=a[i++];
        else
        {
            c[k++]=a[j++];
            cnt+=(m-i+1);
        }
    }
    for(i=0;i<k;i++)
        a[first+i]=c[i];
}
void mergesort(int a[],int first,int last,int c[])
{
    if(first<last)
    {
        int mid=(first+last)/2;
        mergesort(a,first,mid,c);//左边有序 
        mergesort(a,mid+1,last,c);//右边有序 
        merge(a,first,mid,last,c);//再将二个有序数列合并 
    }
}
int main()
{
    int n;
    while(~scanf("%d%I64d",&n,&k))
    {
        memset(c,0,sizeof(c));
        cnt=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        mergesort(a,0,n-1,c);
        if(k>=cnt) cnt=0;
        else cnt-=k;
        printf("%I64d
",cnt);
    }
    return 0;
}