HDU 4911 Inversion（求逆序对）

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 12 2 13 02 2 1

Sample Output

12

题解：

题意：

给你n个数，允许相邻的交换k次，问你能得到的最小逆序对数

思路：

一开始想用树状数组来求逆序对。。后来发现1e9就放弃了。。老老实实归并排序吧，max(逆序对数减去交换次数,0)就是答案，注意要long long

代码：

#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>using namespace std;#define INF 100861111#define ll long long#define eps 1e-15int a[100005];int t[100005];ll ans;void Merge(int l,int r,int mid){    int i=l,j=mid+1,k=l;    while(i<=mid&&j<=r)    {        if(a[i]>a[j])        {            ans+=(mid+1-i);            t[k]=a[j];            j++;        }        else        {            t[k]=a[i];            i++;        }        k++;    }    while(i<=mid)    {        t[k]=a[i];        i++;        k++;    }    while(j<=r)    {        t[k]=a[j];        k++;        j++;    }    for(i=l;i<=r;i++)        a[i]=t[i];}void guibin(int l,int r){    if(l<r)    {        int mid=(l+r)/2;        guibin(l,mid);        guibin(mid+1,r);        Merge(l,r,mid);    }}int main(){    int i,j,n,x;    ll k;    while(scanf("%d%lld",&n,&k)!=EOF)    {        ans=0;        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        guibin(0,n-1);        ans=max(ans-k,(ll)0);        printf("%lld\n",ans);    }    return 0;}