Ultra-QuickSort

最新推荐文章于 2021-09-26 09:12:09 发布

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最新推荐文章于 2021-09-26 09:12:09 发布

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本文介绍了一种名为Ultra-QuickSort的排序算法，并通过具体的代码示例展示了如何计算输入序列进行升序排列所需的最少交换次数。该算法适用于一系列不同的整数，通过递归地将序列分成更小的部分来实现排序。

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 57710		Accepted: 21327

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

Source

Waterloo local 2005.02.05
求交换的次数，归并排序

#include<iostream>
using namespace std;
const int N = 500050;
long long int ans;
void merge(int a[],int l,int mid,int r)
{
    for(int i = l;i < r;i++)
    {
        cout << a[i] << " ";
    }
    cout << endl;
    cout << mid << endl;
    int temp[N];
    int i,j,p = 0;
    i = l,j = mid+1;
    while(i <= mid && j <= r)
    {
        if(a[i] <= a[j])
        {
            temp[p++] = a[i++];
        }
        else
        {
            temp[p++] = a[j++];
            ans = ans +(mid - i + 1);
        }
    }
    while(i <=  mid) temp[p++] = a[i++];
    while(j <= r) temp[p++] = a[j++];
    for(int i = l,p = 0;i <= r;i++)
    {
        a[i] = temp[p++];
    }
}
void mergesort(int a[],int l,int r)
{
    if(l == r) a[r] = a[r];
    else
    {
        int mid = (l + r)/2;
        mergesort(a,l,mid);
        mergesort(a,mid+1,r);
        merge(a,l,mid,r);
    }
}
int main()
{
    int n,a[N];
    while(cin >> n,n)
    {
        ans = 0;
        for(int i = 0;i < n;i++)
        {
            cin >> a[i];
        }
        mergesort(a,0,n-1);
        cout << ans << endl;
    }
    return 0;
}