HDU 2689Sort it 树状数组 逆序对

Posted 2020-08-12 GeekZRF

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4110    Accepted Submission(s): 2920

Problem Description

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

 

 

Output

For each case, output the minimum times need to sort it in ascending order on a single line.

 

 

Sample Input

3

1 2 3

4

4 3 2 1

 

Sample Output

0

6

 

 

Author

WhereIsHeroFrom

 

 

Source

ZJFC 2009-3 Programming Contest

 

 

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题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=2689

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题意：交换相邻的两个数，使得序列是上升的。问需要交换多少次。

思路：求出每个数ai前面有多少个数比ai大，或者每个数ai后面有多少个数比ai小。第一种方法只需要交换树状数组更新和求和的函数。第二种方法就是求逆向对的数量和只需要逆向输入a，直接标准的树状数组。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=1100;
int c[MAXN];
inline int Lowbit(int x)
{
    return x&(-x);
}
void update(int i,int val)
{
    for(i; i>0; i-=Lowbit(i))
        c[i]+=val;
}
int sum(int i)
{
    int temp=0;
    for(i; i<=MAXN; i+=Lowbit(i))
        temp+=c[i];
    return temp;
}
int main()
{
    int i,n;
    int a;
    while(~scanf("%d",&n))
    {
        int ans=0;
        memset(c,0,sizeof(c));
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a);
            ans+=sum(a);
            update(a,1);
        }
        cout<<ans<<endl;
    }
    return 0;
}

求左边大于等于 a[i]的数的个 数