HOJ2098//归并排序求逆系数

最新推荐文章于 2021-12-06 00:09:43 发布

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最新推荐文章于 2021-12-06 00:09:43 发布

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分类专栏： Data structure and Algorithm 文章标签： input integer sorting merge output algorithm

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本文探讨了Ultra-QuickSort算法处理特定整数序列时所需的交换操作数量，并通过归并排序方法计算逆序数，适用于序列排序效率分析。

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Ultra-QuickSort

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	Source : Waterloo ACM Programming Contest Feb 5, 2005
	Time limit : 3 sec		Memory limit : 32 M

Submitted : 519, Accepted : 171

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<iostream>
#include<stdio.h>
#define MAX  500000
using namespace std;
int a[MAX +1],t[MAX +1];
long long nxs;
void Merge(int left, int mid, int right)//这个是照数据结构上面的模版写的，感觉写的很清晰,以后就用它做模版啦
{
    int i = left, j = mid + 1,k=0;
    while(i <= mid && j <= right)
    {
        if (a[i] > a[j])
        {
            t[k++] = a[j++];
            nxs += mid - i +1;// a[i]后面的数字对于a[j]都是逆序的
        }
        else
        {
            t[k++] = a[i++];
        }
    }
    while(i <= mid) t[k++] = a[i++];
    while(j <= right) t[k++] = a[j++];
    for (i = 0; i < k; i++)//这一步是必须的，归并完成后将结果复制到原输入数组
    {
      a[left+i]=t[i];
    }
}
void MergeSort(int left, int right)
{
    int mid;
    if (left < right)
    {
        mid = (left + right) / 2;
        MergeSort(left, mid);
        MergeSort(mid + 1, right);
        Merge(left, mid, right);
    }
}
int main()
{
    int i,n;
    while(scanf("%d", &n)&&n)
    {
        nxs=0;
        for(i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
        }
        MergeSort(0, n - 1);
        printf("%lld\n", nxs);
    }
    return 0;
}

主要就是利用归并排序求逆系数，以前没有学过排序，思路都是别人的，当作学习吧，嘻嘻！