poj 2299 Ultra-QuickSort (归并求逆序数)

最新推荐文章于 2021-10-09 14:23:30 发布

K_y

最新推荐文章于 2021-10-09 14:23:30 发布

阅读量278

点赞数

分类专栏：归并 poj 文章标签： poj 归并

本文链接：https://blog.youkuaiyun.com/ky961221/article/details/52049496

版权

poj 同时被 2 个专栏收录

53 篇文章

订阅专栏

归并

1 篇文章

订阅专栏

本文介绍了一种名为Ultra-QuickSort的排序算法，该算法通过相邻元素交换实现序列升序排列。针对大规模数据集，文章提供了一种高效的归并排序实现方式，并附带AC代码，用于计算将无序序列变为有序所需交换次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 54525		Accepted: 20039

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

题意：给你一串无序的数，让你求把他变成有序的，需要交换的次数

ｎ比较大，冒泡，暴力找都会超时

So,现学的归并排序

归并排序模板

void merge_sort(long long a[],int x,int y,long long t[])　　／／ｔ是临时数组
{
    if(y-x>1)　／／如果分到还剩一个数，递归停止
    {
        int m=x+(y-x)/2;
        int p=x,q=m,i=x;
        merge_sort(a,x,m,t);／／右边有序，一直分到还剩一个
        merge_sort(a,m,y,t);／／左边有序．．．
        while(p<m || q<y)　／／合并
        {
            if(q>=y || (p<m && a[p]<=a[q]))　／／左边没有数了或者右边有数但右边的数小于左边的数
            {
                t[i++]=a[p++];
            }
            else
            {
                t[i++]=a[q++];
                ct+=m-p;／／求逆序对（如果右边的大于左边的，就是逆序）
            }
        }
        for(i=x;i<y;i++)
            a[i]=t[i];／／整合到源数组
    }

}

AC代码（注意long long）

#include <stdio.h>
long long a[512345],t[512345];
long long ct;
void merge_sort(long long a[],int x,int y,long long t[])
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;
        int p=x,q=m,i=x;
        merge_sort(a,x,m,t);
        merge_sort(a,m,y,t);
        while(p<m || q<y)
        {
            if(q>=y || (p<m && a[p]<=a[q]))
            {
                t[i++]=a[p++];
            }
            else
            {
                t[i++]=a[q++];
                ct+=m-p;
            }
        }
        for(i=x;i<y;i++)
            a[i]=t[i];
    }

}
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        ct=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
        }
        merge_sort(a,0,n,t);
        printf("%lld\n",ct);

    }
}