hdu 4911 Inversion

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 68    Accepted Submission(s): 23

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

树状数组求逆序对，如果总对数大于k 就减k，否则置为0

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
#include<cstring>
#include<stdlib.h>
#include<string>
#include<cmath>
#include<map>
using namespace std;
#define pb push_back
#define mmax 100000
#define ll __int64
#define mod 1000000007
ll p[101000];
int ac[101000],tmp[101000],cnt;
int n,m;
int Find(int x){
    int l=1,r=cnt,mid,tt=10000000;
    while(l<=r){
        mid=(l+r)>>1;
        if(tmp[mid]>=x) tt=min(tt,mid),r=mid-1;
        else l=mid+1;
    }
    return tt;
}
void update(int pos,int num){
    while(pos>0){
        p[pos]+=num;
        pos-=pos&(-pos);
    }
}
ll getnum(int pos){
    ll sum=0;
    while(pos<=cnt){
        sum+=p[pos];
        pos+=pos&(-pos);
    }
    return sum;
}
int main(){
    while(cin>>n>>m){
        for(int i=1;i<=n;i++){
             scanf("%d",&ac[i]);
             tmp[i]=ac[i];
        }
        memset(p,0,sizeof(p));
        sort(tmp+1,tmp+1+n);离散化
        cnt=0;
        tmp[++cnt]=tmp[1];
        ll sum=0;
        for(int i=2;i<=n;i++) if(tmp[i]!=tmp[cnt]) tmp[++cnt]=tmp[i];
        for(int i=1;i<=n;i++){
            int x=Find(ac[i]);
            sum+=getnum(x+1);
            update(x,1);
        }
        if(sum>=m) sum-=m;
        else sum=0;
        printf("%I64d
",sum);
    }
}