POJ训练计划2299_Ultra-QuickSort(归并排序求逆序数)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 39279		Accepted: 14163

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

解题报告

求相邻的两两交换排序要几个步骤，就是求逆序数。

求逆序数的n方算法肯定超时，网上介绍了高速求逆序数的算法是用归并排序来实现的，时间复杂度是O（nlogn）。

刚学习了归并排序，这里不介绍归并排序。

为什么归并排序能够求逆序数。

在一个排列中，假设一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数。依据归并排序，“划分”把序列分成元素个数尽量相等的两半，“递归”统计i和j均在左边或是均在右边的逆序对个数；“合并”统计i在左边，但j在右边的逆序数个数。

怎么统计逆序数个数呢？在归并排序合并操作时，因为是从小到大的排序，当A[j]复杂到T中时，左边还没有来得及复制的那些数就是左边全部比A[j]大的数。（以上i表示左区间数组的指针，j表示右区间数组指针，T是辅助空间，A是原数组）

这个题目要注意的是复杂空间必须开全局，局部可能溢出，还有答案要用longlong存。

#include <iostream>

using namespace std;
long long a[500010],t[500010],cnt=0;
void gbsort(long long *a,long long l,long long r)
{
    if(l<r)
    {
        long long mid=(l+r)/2;
        gbsort(a,l,mid);
        gbsort(a,mid+1,r);
        long long s=l,e=mid+1;
        long p=0;

        while(s<=mid&&e<=r)
        {
            if(a[s]<=a[e])
            {
                t[p++]=a[s++];
            }
            else
            {
                t[p++]=a[e++];
                cnt+=mid-s+1;
            }
        }
        while(s<=mid)
        {
            t[p++]=a[s++];
        }
        while(e<=l)
        {
            t[p++]=a[e++];
        }
        for(int i=0;i<p;i++)
            a[l+i]=t[i];
    }
}
int main()
{
    int n;
    while(cin>>n)
    {
        if(!n)break;
        cnt=0;
        for(int i=0; i<n; i++)
            cin>>a[i];
        gbsort(a,0,n-1);
        cout<<cnt<<endl;
    }
    return 0;
}