poj 2299 Ultra-QuickSort

最新推荐文章于 2020-03-19 20:17:52 发布

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本文介绍了一种名为Ultra-QuickSort的排序算法，该算法通过计算输入序列的逆序数来确定排序所需的最少交换次数。文章提供了一个C++程序示例，使用归并排序策略来高效地解决这一问题。

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题目链接：点击打开链接

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

题目大意：给数一数字序列，按递增排序，两个相邻的做交换，交换几次

ps：一看就是冒泡排序的步骤，但是会超时，；

这里用归并排序求逆序数（离散中讲过），逆序数就是它的交换次数（不要问为什么，模拟几个例子就知道了）

<span style="font-size:18px;">#include <iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
int a[500100];
int aux[500100];
long long ans;
void compute(int a[],int l,int mid,int r)
{
    for(int i=l;i<=r;i++)
    {
        aux[i]=a[i];
    }
    int i=l;
    int j=mid+1;
    for(int k=l;k<=r;k++)
    {
        if(i>mid)a[k]=aux[j++];
        else if(j>r)a[k]=aux[i++];
        else if(aux[i]<aux[j])a[k]=aux[i++];
        else
        {
            a[k]=aux[j++];
            ans+=mid-i+1;
        }
    }
}
void mergesort(int a[],int l,int r)
{
    if(l<r)
    {
        int mid=(l+r)/2;
        mergesort(a,l,mid);
        mergesort(a,mid+1,r);
        compute(a,l,mid,r);
    }
}
int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        ans=0;
        mergesort(a,1,n);
        printf("%lld\n",ans);
    }
    return 0;
}
</span>