POJ-2299（树状数组）

最新推荐文章于 2019-09-16 22:48:47 发布

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本文深入探讨了Ultra-QuickSort排序算法的工作原理，详细解释了如何通过交换相邻元素将序列排序为升序的过程。文章提供了算法的具体实现，包括输入序列的处理、排序操作计数及输出结果的方法。此外，还介绍了使用树状数组进行前缀和查询以优化算法效率的技术。

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 75430		Accepted: 28242

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

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
    int w,i;
};
int a[500001],c[500001],n;
node x[500001];

bool cmp(node x,node y)
{
    return x.w<y.w;
}

int lowbit(int x)
{
    return x&(-x);
}

int sum(int pos)
{
    int s=0;
    while(pos>0)
    {
        s+=c[pos];
        pos-=lowbit(pos);
    }
    return s;
}

void update(int pos,int num)
{
    while(pos<=n)
    {
        c[pos]+=num;
        pos+=lowbit(pos);
    }
}

int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x[i].w);
            x[i].i=i;
        }

        sort(x+1,x+1+n,cmp);
        for(int i=1;i<=n;i++)
            a[x[i].i]=i;

        memset(c,0,sizeof(c));
        long long ans=0;
        for(int i=1;i<=n;i++)
        {
            update(a[i],1);
            ans+=i-sum(a[i]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}