Ultra-QuickSort【归并排序典型题目】

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 34470		Accepted: 12382

题目链接：http://poj.org/problem?id=2299

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

 

题目大意：给出一段数字序列，每次只能交换相邻的两个数，问最少需要多少次可以使得这段序列变成顺序序列。

思路：用冒泡排序超时，因为给的数据量很大。用归并排序的方法可以求出逆序数，在这里，逆序数就是冒泡排序的交换次数，所以这道题目就是用归并排序的方法对数组进行排序，并得出逆序数，输出即可。

代码：

#include<iostream>
#include<string.h>
#include<cstdio>
#include<cstdlib>
#define max 1000000
using namespace std;
int n,f[max],g[max];
long long int sum=0;
void guibing(int l,int mid,int r)
{
    int t=0;
    int i=l,j=mid+1;
    while(i<=mid&&j<=r)
    {
        if(f[i]<=f[j])
        {
            g[t++]=f[i];
            i++;
        }
        else
        {
                g[t++]=f[j];
                j++;
                sum=sum+mid-i+1;
        }
    }
    while(i<=mid)g[t++]=f[i++];
    while(j<=r)g[t++]=f[j++];
    for(i=0;i<t;i++)
    f[l+i]=g[i];
}
void quicksort(int l,int r)
{
    if(l<r)
    {
        int mid=(l+r)/2;
        quicksort(l,mid);
        quicksort(mid+1,r);
        guibing(l,mid,r);
    }
}
int main()
{
    while(1)
    {
        cin>>n;
        if(n==0)break;
        int i;
        sum=0;
        for(i=0;i<=n-1;i++)
        cin>>f[i];
        quicksort(0,n-1);
        cout<<sum<<endl;
    }
    return 0;
}