poj2299.4thweek.p.A.归并排序

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
分析：
1.划分问题 2.递归求解. 3.合并
代码及步骤分析：

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=500005;
int a[maxn],b[maxn];
long long k;

int merge_sort(long long x,long long y)
{
    if(x<y)
    {
        long long m=(x+y)/2;   //划分区间
        merge_sort(x,m);       //划分并递归求解
        merge_sort(m+1,y); 
        int p=0,i=x,j=m+1;
        while(i<=m && j<=y)    //划分的数组两边都非空时
        {
            if(a[i]>a[j])     //找到划分后的数组左边的数大于右边的数时
            {
                b[p++]=a[j++];//  将左边数组复制到临时空间
                k+=m+1-i;      //注意此时k值的记录！！！
            }
            else
                b[p++]=a[i++];   //将右边数组复制到临时空间
            
        }
        while(i<=m)
            b[p++]=a[i++];   
        while(j<=y)
            b[p++]=a[j++];
        for(i=0;i<p;++i)
            a[x+i]=b[i];
        
    }
    return 0;
}

int main()
{
    long long n,i;
    while(scanf("%lld",&n) && n)
    {   
        k=0;  //定义k记录交换次数
        for(i=0;i<n;i++)
            scanf("%lld",&a[i]);
        merge_sort(0,n-1);   
        printf("%lld
",k);
    }
    
    return 0;
}

归并排序时间复杂度为O（nlogn），统计改变次数k也并不改变其时间复杂度。