POJ 2299 Ultra-QuickSort（逆序数、分治）

最新推荐文章于 2021-01-24 14:23:50 发布

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最新推荐文章于 2021-01-24 14:23:50 发布

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分类专栏：分治

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本文深入探讨了Ultra-QuickSort算法的原理与实现细节，详细解释了如何通过统计逆序数来确定序列排序所需的最少交换次数，并提供了代码实现，旨在帮助读者理解排序算法的基本概念与实践应用。

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 43803		Accepted: 15974

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

解题报告：根据题意是求交换相邻两位至少需要多少次才能让数组升序。其实就是求逆序数，求逆序数其实和归并排序是一样的，就是在归并的过程中，加入逆序数的统计。例如现在有两个升序数组1 4 9 和 2 6 7 ，那么在归并过程中，插入前一个数组不是逆序，而一旦插入后一个数组中元素，2那么对应的逆序数就加2 。 (因为4和9都还没有插入。)

/********************************/
/*Problem:		POJ 2299	    */
/*User: 	    shinelin	    */
/*Memory: 	    7228K		    */
/*Time: 	    438MS		    */
/*Language: 	C++		        */
/********************************/
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <string>
#include <list>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

#define INF 0x7fffffff
#define LL long long
#define MAXN 500005

LL a[MAXN], com[MAXN];
LL cnt = 0;

void swapTimes(int low, int high)
{
    if(low == high - 1)
        return ;
    if(low == high - 2)
    {
        if(a[low] > a[low + 1])
        {
            swap(a[low], a[low + 1]);
            cnt ++;
        }
        return ;
    }
    int mid = (low + high) >> 1;
    swapTimes(low, mid);
    swapTimes(mid, high);
    int i = low, j = mid;
    int k = low;    //三个游标
    while (k < high)
    {
        if(j == high|| i < mid && a[i] <= a[j])
        {
            com[k++] = a[i++];
        }

        if(k == high) break;
        if(i == mid || j < high && a[i] > a[j])
        {
            com[k++] = a[j++];
            cnt += mid - i;     //归并排序的过程中，统计交换次数。
        }
    }
    for (int i = low; i < high; i ++)
    {
        a[i] = com[i];
    }
    return ;
}
int main()
{
    int n;
    while(~scanf("%d", &n), n)
    {
        cnt = 0;
        for (int i = 0; i < n; i ++)
        {
            scanf("%I64d", a + i);
        }
        swapTimes(0, n);
        printf("%I64d\n", cnt);
//        for (int i = 0; i < n; i ++)
//        {
//            printf("%d ", a[i]);
//        }
//        printf("\n");
    }
    return 0;
}