ACM: 计算交换次数的排序题 poj 22…

                                                                 Ultra-QuickSort

Description

In this problem, you have toanalyze a particular sorting algorithm. The algorithm processes asequence of n distinct integers by swapping two adjacent sequenceelements until the sequence is sorted in ascending order. For theinput 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-QuickSortneeds to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case beginswith a line that contains a single integer n <500,000 -- the length of the input sequence. Each of the thefollowing n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999,the i-th input sequence element. Input is terminated by a sequenceof length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single linecontaining an integer number op, the minimum number of swapoperations necessary to sort the given input sequence.

Sample Input

5

9 1 0 5 4

3

1 2 3

0

Sample Output

6

0

题意: 将序列排序,从小到大.计算交换的次数,交换只可以和左右相邻的数字交换.

 

解题思路:

               1. 交换排序的类型,选择归并排序.

               2. 计算二路归并的时候, mid - p + 1 (p: 序列的下标,mid: 是二分的中介.)

 

代码:

#include <cstdio>
#include <iostream>
using namespace std;
#define MAX 500003

int a[MAX];
int t[MAX];
int n;

long long merge_sort(int start,int end,int len)
{
    if(len ==0)
      return 0;
    long longcur = 0;
    int mid =(start+end) / 2;
    cur +=merge_sort(start,mid,mid-start);
    cur +=merge_sort(mid+1,end,end-mid-1);
    int p =start , q = mid+1;
    int k =start;
    
    for(int i =0; i <= len; ++i)
    {
      if(q > end || (p <=mid && a[p] <a[q]))
         t[k++] = a[p++];
      else
      {
         cur += mid-p+1;
         t[k++] = a[q++];
      }
    }
    for(;start<= end; ++start)
      a[start] = t[start];

    returncur;
}

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d",&n) != EOF&& n != 0)
    {
      for(int i = 0; i < n; ++i)
      {
         scanf("%d",&a[i]);
      }
      printf("%lld\n",merge_sort(0,n-1,n-1));
    }
    return0;
}