HDU 4911 inversion (merge sort to calculate the number of reverse orders)

Merge and sort to obtain the number of reverse orders, and then obtain the maximum value between ANS-K and 0.

You can also use a tree array. During the game, the posture may be incorrect. The tree array wa ,,

Inversion Time Limit: 2000/1000 MS (Java/others) memory limit: 131072/131072 K (Java/Others)
Total submission (s): 578 accepted submission (s): 249


Problem descriptionbobo has a sequence A1, A2 ,..., An. He is allowed to swap two AdjacentNumbers 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 AI> AJ.
Inputthe 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 ,..., An (0 ≤ AI ≤ 109 ).
Outputfor each tests:

A single integer denotes the minimum number of inversions.
Sample Input

3 12 2 13 02 2 1

 
Sample output

12

 

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-12///#define M 1000100#define LL __int64///#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 10010000;int num[maxn];LL ans;int tmp[maxn];void Sell(int l, int mid, int r){    int i = l;    int j = mid+1;    int k = 0;    while(i <= mid && j <= r)    {        if(num[i] > num[j])        {            tmp[k++] = num[j++];            ans += mid-i+1;        }        else tmp[k++] = num[i++];    }    while(i <= mid)        tmp[k++] = num[i++];    while(j <= r)        tmp[k++] = num[j++];    for(int i = 0; i < k; i++)        num[i+l] = tmp[i];}void Sort(int l, int r){    if(l < r)    {        int mid = (l+r)/2;        Sort(l, mid);        Sort(mid+1, r);        Sell(l, mid, r);    }}int main(){    int n, k;    while(cin >>n>>k)    {        ans = 0;        for(int  i = 0;i < n; i++) scanf("%d",&num[i]);        Sort(0, n-1);        cout<<max(0LL, ans-k)<<endl;    }    return 0;}