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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.
Ultra-QuickSort produces the output
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; }
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