Ultra-QuickSort——［归并排序、分治求逆序对］

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
解题思路：
　　本题揭示了排序的本质——消除逆序对。逆序对，是指对于满足 当i<j时，a[i]>a[j]的序偶（a[i],a[j]）。可以证明，交换序列中任意两个
相邻元素，逆序对增加或减少一。那么对于一个给定的序列，按一次交换一对相邻元素的方法，最少需要要交换的次数等于此序列中逆序对的数目。
那么问题就变成了如何求给定序列的逆序对。可以采用分治的方法：
　　整个序列逆序对数＝左半序列逆序对数＋右半序列逆序对数＋前一个元素在左半序列，后一个元素在右半序列的逆序对数。
　　再细考虑可以发现，这个过程可以在归并排序的同时完成，时间复杂度为O(NlogN).

注意：可以简单计算一下，n个元素的排列逆序对数最多有 n（n－1）／2个，需要用到long long。

代码如下:

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <ctime>
 using namespace std;
 #define print_time_ printf("time : %f\n",double(clock())/CLOCKS_PER_SEC)
 #define maxn 500000
 int A[maxn+];

 typedef long long LL;
 LL DC(int a,int N){
     int mid=a+N/-;
     int b=a+N-;
     if(N==)
         return ;

     LL x1=DC(a, N/);
     LL x2=DC(mid+,b-mid);
     LL x3=;
     int *B=new int[N];
     int i=a,j=mid+,p=;
     for(;i<=mid&&j<=b&&p<N;p++){
         if(A[i]<=A[j]){
             B[p]=A[i++];
         }
         else {
             B[p]=A[j++];
             x3+=mid-i+;
         }
     }
     while(i<=mid){B[p++]=A[i++];}
     while(j<=b){B[p++]=A[j++];}
     memcpy(A+a, B, N*sizeof(int));
     delete [] B;
     return x1+x2+x3;
 }
 int main() {
     int n;
     while(scanf("%d",&n)==&&n){
         for(int i=;i<n;i++)
             scanf("%d",&A[i]);
         printf("%lld\n",DC(, n));
     }
     //print_time_;
     return ;
 }