HDU4911-Inversion(树状数组)

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 914    Accepted Submission(s): 380

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

 

题意：n个数。最多有k次相邻位置的数的交换，问最小的逆序数为多少

思路：保证每次交换逆序数都减小，能够參考冒泡排序的做法。最后仅仅要计算原数列逆序数，然后减掉k就可以。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int maxn = 100000+10;
int sum[maxn];
int n,k;
int num[maxn],tn[maxn];
map<int,int> mma;
bool cmp(int a,int b){
    return a > b;
}
int lowbit(int x){
    return x&(-x);
}
void add(int x,int d){
    while(x < maxn){
        sum[x] += d;
        x += lowbit(x);
    }
}
int getS(int x){
    int ret = 0;
    while(x > 0){
        ret += sum[x];
        x -= lowbit(x);
    }
    return ret;
}
int main(){

    while(~scanf("%d%d",&n,&k)){
        mma.clear();
        memset(sum,0,sizeof sum);
        for(int i = 1; i <= n; i++){
            scanf("%d",&num[i]);
            tn[i] = num[i];
        }
        sort(tn+1,tn+n+1,cmp);
        int i = 1,cnt = 1;
        while(i <= n){
            mma[tn[i]] = cnt;
            int j = i+1;
            while(j <= n && tn[i]==tn[j]){
                j++;
            }
            cnt++;
            i = j;
        }
        long long ret = 0;
        for(int i = 1; i <= n; i++){
            ret += getS(mma[num[i]]-1);
            add(mma[num[i]],1);
        }
        long long  ans = ret-k;
        if(ans < 0) ans = 0;
        cout<<ans<<endl;
    }
    return 0;
}