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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,网上介绍了快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。
为什么归并排序可以求逆序数。
在一个排列中,如果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。根据归并排序,“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,由于是从小到大的排序,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所有比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;
}