hdu 4911 Inversion （ 2014 Multi-University Training Contest 5）

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1451    Accepted Submission(s): 591

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

 

题解：

       根据定理，存在每次相邻的两个数交换，使逆序数减一，用归并排序求出逆序数再减交换次数即可。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn=100100;
const int maxint= 999999999;
int a[maxn];
int lef[maxn];
int righ[maxn];
int n;
long long ans;
void sort(int l,int temp,int r)
{
    int n1=temp-l+1;
    int n2=r-temp;
    for(int i=0;i<n1;i++)
    lef[i]=a[i+l];
    for(int i=0;i<n2;i++)
    righ[i]=a[temp+i+1];
    lef[n1]=righ[n2]=maxint;
    int i=0;
    int j=0;
    for(int k=l;k<=r;k++)
    {
        if(lef[i]<=righ[j])
        {
            a[k]=lef[i];
            i++;
        }
        else
        {
            a[k]=righ[j];
            j++;
            ans+=n1-i;//计算逆序数，在lef数组中有n1-i个比righ[j]大的
        }
    }
    return;

}
void memsort(int l,int r)
{
    int temp=(l+r)/2;
    if(l<r)
    {
    memsort(l,temp);
    memsort(temp+1,r);
    sort(l,temp,r);
    }
    return;
}
int main()
{
    int k;
    while(~scanf("%d%d",&n,&k))
    {
        ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        memsort(0,n-1);//归并排序
        if(ans-k<0)
        ans=k;
        printf("%I64d\n",ans-k);
    }
    return 0;
}