hdu-5497 Inversion(滑动窗口+树状数组)

最新推荐文章于 2022-03-09 14:35:36 发布

转载最新推荐文章于 2022-03-09 14:35:36 发布 · 85 阅读

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原文链接：http://www.cnblogs.com/zhangchengc919/p/5922283.html

本文介绍了一种解决特定问题的算法：在一个整数序列中删除长度为m的连续子序列，以使得剩余序列的逆序对数量达到最小。通过使用树状数组和滑动窗口技巧来高效计算逆序对的数量。

题目链接：

Inversion

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1087 Accepted Submission(s): 323

Problem Description

You have a sequence

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test case, output the minimum number of inversions.

Sample Input

3 1

1 2 3

4 2

4 1 3 2

Sample Output

题意：

有一个序列,

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
int sum1[maxn],sum2[maxn],n,m,a[maxn];
int lowbit(int x){return x&(-x);}
inline void update1(int x,int num)
{
    while(x<=n)
    {
        sum1[x]+=num;
        x+=lowbit(x);
    }
}
inline int query1(int x)
{
    int s=0;
    while(x)
    {
        s+=sum1[x];
        x-=lowbit(x);
    }
    return s;
}
inline void update2(int x,int num)
{
    while(x<=n)
    {
        sum2[x]+=num;
        x+=lowbit(x);
    }
} 
inline int query2(int x)
{
    int s=0;
    while(x)
    {
        s+=sum2[x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++)sum1[i]=sum2[i]=0;
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        LL su=0;
        for(int i=n;i>0;i--)
        {
            su=su+query2(a[i]-1);
            update2(a[i],1);
        }
        LL ans=su,temp=su;int l,r=1;
        for(l=1;l<=n-m+1;l++)
        {
            while(r-l<m&&r<=n)
            {
                update2(a[r],-1);
                temp=temp-query2(a[r]-1);
                temp=temp-(l-1-query1(a[r]));
                r++;
            }
            ans=min(ans,temp);
            temp=temp+(l-1-query1(a[l]));
            temp=temp+query2(a[l]-1);
            update1(a[l],1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/zhangchengc919/p/5922283.html