HDU 4911 Inversion【归并排序求逆序数】

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2725    Accepted Submission(s): 1008

Problem Description

bobo has a sequence a₁,a₂,…,a_n. He is allowed to swap twoadjacent 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

嗯，题目大意就是说，n和k，n代表数据个数，k代表数据相邻之间交换的次数限制。还是归并排序找到排到从小到大的最小次数然后减去k 就是最小的inversions（因为归并排序是稳定的，而sort不是稳定的）（这里我实验证明用malloc多次申请内存再释放是很费时间的，在主函数里申请一次就ok这里TLE了我几次，然后还是比直接申请一个数组要节省一点点时间的，节省了大概80ms）

 

#include <iostream>#include<cstdlib>#include<cstdio>#include<cstring>#define maxn 100000+10using namespace std;long long n,cnt,k,a[maxn],p[maxn];void Merge(long long left,long long mid,long long right,long long *p){    long long n=mid,m=right;    long long i=left,j=mid+1;    long long num=0;    while(i<=n&&j<=m)    {        if(a[i]<=a[j])            p[num++]=a[i++];        else        {            p[num++]=a[j++];            cnt+=n-i+1;        }    }    while(i<=n)        p[num++]=a[i++];    while(j<=m)        p[num++]=a[j++];    for(int i=0;i<num;++i)        a[left+i]=p[i];}void mergesort(long long left,long long right,long long *p){    if(left<right)    {        long long mid=(left+right)/2;        mergesort(left,mid,p);        mergesort(mid+1,right,p);        Merge(left,mid,right,p);    }}int main(){    while(cin>>n>>k)    {        for(int i=0;i<n;++i)            cin>>a[i];        cnt=0;        long long *p=(long long *)malloc(sizeof(long long)*n);        mergesort(0,n-1,p);        if(cnt<=k)            cout<<0<<endl;        else            cout<<cnt-k<<endl;        free(p);    }    return 0;}