poj2299 Ultra-QuickSort 树状数组-优快云博客

本文介绍了一种使用树状数组优化Ultra-QuickSort算法中逆序对计数的方法。通过两次排序和树状数组更新及查询操作，实现了高效地计算输入序列排序所需的最小交换次数。

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Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

Sample Output

6
0

求逆序对用树状数组这样每次是 log(n)的更新

一定要理解树状数组的过程直接拍板是不好的具体的树状数组的实现我会另转载一个文章，心得蛮不错的，理解之后会觉得非常的这个算法很有效。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=500050;
ll n;
ll c[N];
struct node
{
    ll order,dig;
}a[N];
ll lowbit(ll x)
{
   return (x&(-x));
}
ll calc(ll x)
{
    ll sum=0;
    for(;x;x-=lowbit(x))
    {
        sum+=c[x];
    }
    return sum;
}
void update(ll x,ll val)
{
    while(x<=n)
    {
        c[x]+=val;
        x+=lowbit(x);
    }
}
bool cmp1(const node &a,const node &b)
{
    return a.dig<b.dig;
}
bool cmp2(const node &a,const node &b)
{
    return a.order<b.order;
}
int main()
{
    while(scanf("%lld",&n)&&n)
    {
        ll sum=0;
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i].dig);
            a[i].order=i;
        }
        sort(a+1,a+n+1,cmp1);
        for(int i=1;i<=n;i++)
        {
            a[i].dig=i;
        }
        sort(a+1,a+n+1,cmp2);
        //离散完毕
        for(int i=1;i<=n;i++)
        {
            update(a[i].dig,1);
            sum+=i-calc(a[i].dig);
        }
        printf("%lld\n",sum);
    }
    return 0;
}