HDU 4911 Inversion（归并排序求逆序对）

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 12 2 13 02 2 1

 

Sample Output

12

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 【题解】

  水题，不过题意比较模糊，给一个序列，每次可以交换相邻的逆序数（i<j&&a[i]?a[j]），

 并且最多交换k次，问交换完的最小逆序对个数。

 思考一下，每次只交换相邻两个，所以也就是说每次只会减少一个逆序对，所以就是把最初的逆序数对求出来减去k即为答案。

 注意要用  long long

 【AC代码】

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int N=1e5+5;int n,node[N],m[N],str[N];__int64 ss,mins;void Mergee(int first,int mid,int end,int a[]){    int i=first;    int j=mid+1;    int k=first;    while(i<=mid && j<=end)    {        if(a[i]>a[j])        {            m[k++]=a[j++];            ss += mid - i+1 ;        }        else        {            m[k++]=a[i++];        }    }    while(i<=mid)    {        m[k++]=a[i++];    }    while(j<=end)    {        m[k++]=a[j++];    }    for(int i=first;i<=end;i++)        a[i]=m[i];}void MergeSort(int first,int end,int a[]){    if(first<end)    {        int mid = (first+end)>>1;        MergeSort(first,mid,a);        MergeSort(mid+1,end,a);        Mergee(first,mid,end,a);    }}int main(){    int k;     while(~scanf("%d%d",&n,&k))     {         for(int i=1;i<=n;i++)         {             scanf("%d",&node[i]);         }         ss=0;         MergeSort(1,n,node);         mins=ss;         printf("%I64d\n",max(mins-k,(__int64)0));     }     return 0;}