POJ 2299 Ultra-QuickSort (归并排序求逆序数)

Ultra-QuickSort

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

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

Source

Waterloo local 2005.02.05

题目链接：http://poj.org/problem?id=2299

题目大意：求逆序数

题目分析：以前学的用树状数组求逆序数，补一下归并排序的求法，感觉实现起来更简单，归并排序自顶向下分解，自底向上合并，每次合并的两个区间都是已经排好序了的，这就给我们求逆序数带来了很大的好处

我们把一个大区间[l,r]分成[l,mid], [mid + 1, r]，显然每次我们只要求一个数在左区间，一个数在右区间时的逆序数个数，而不用考虑左区间内和右区间内的逆序数个数，因为合并是自底向上的，左区间和右区间内的逆序数我们已经在他们的子状态中求结果了，所以在自底向上合并时，我们直接累加每一层的逆序数个数就是最后整个区间的逆序数了。很赞的应用，对递归有了更深刻的理解

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 500005;
int a[MAX], n;
ll ans;

void Merge(int l, int mid, int r)
{
    int i = l, j = mid + 1;
    while(i <= mid && j <= r)
    {
        if(a[i] <= a[j])
            i ++;
        else
        {
            j ++;
            //因为左右区间都是有序的，因此a[i]>a[j]说明a[i]~a[mid]都大于a[j]
            ans += mid - i + 1;
        }
    }
    sort(a + l, a + r + 1);
    return;
}

void Merge_sort(int l, int r)
{
    if(l < r)
    {
        int mid = (l + r) / 2;
        Merge_sort(l, mid);
        Merge_sort(mid + 1, r);
        Merge(l, mid, r);
    }
    return;
}

int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        ans = 0;
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        Merge_sort(0, n - 1);
        printf("%lld\n", ans);
    }
}

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