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:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <ctime>
 5 using namespace std;
 6 #define print_time_ printf("time : %f\n",double(clock())/CLOCKS_PER_SEC)
 7 #define maxn 500000
 8 int A[maxn+5];
 9
10 typedef long long LL;
11 LL DC(int a,int N){
12     int mid=a+N/2-1;
13     int b=a+N-1;
14     if(N==1)
15         return 0;
16
17     LL x1=DC(a, N/2);
18     LL x2=DC(mid+1,b-mid);
19     LL x3=0;
20     int *B=new int[N];
21     int i=a,j=mid+1,p=0;
22     for(;i<=mid&&j<=b&&p<N;p++){
23         if(A[i]<=A[j]){
24             B[p]=A[i++];
25         }
26         else {
27             B[p]=A[j++];
28             x3+=mid-i+1;
29         }
30     }
31     while(i<=mid){B[p++]=A[i++];}
32     while(j<=b){B[p++]=A[j++];}
33     memcpy(A+a, B, N*sizeof(int));
34     delete [] B;
35     return x1+x2+x3;
36 }
37 int main() {
38     int n;
39     while(scanf("%d",&n)==1&&n){
40         for(int i=0;i<n;i++)
41             scanf("%d",&A[i]);
42         printf("%lld\n",DC(0, n));
43     }
44     //print_time_;
45     return 0;
46 }