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=;
 void guibing(int l,int mid,int r)
 {
     int t=;
     int i=l,j=mid+;
     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+;
         }
     }
     while(i<=mid)g[t++]=f[i++];
     while(j<=r)g[t++]=f[j++];
     for(i=;i<t;i++)
     f[l+i]=g[i];
 }
 void quicksort(int l,int r)
 {
     if(l<r)
     {
         int mid=(l+r)/;
         quicksort(l,mid);
         quicksort(mid+,r);
         guibing(l,mid,r);
     }
 }
 int main()
 {
     while()
     {
         cin>>n;
         if(n==)break;
         int i;
         sum=;
         for(i=;i<=n-;i++)
         cin>>f[i];
         quicksort(,n-);
         cout<<sum<<endl;
     }
     return ;
 }