NUC_TeamTEST_C && POJ2299（只有归并）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 42627		Accepted: 15507

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

这道题就是通过求逆序数的和，来求出序列所有的交换次数。用冒泡排序直接会TLE（相当于暴力了），

这道题有3种解法，树状数组、线段树、归并排序（O（N*lgN））

其中归并排序的写法应该是最简单的，树状数组、线段树要用到离散化，博客后续还会跟上

归并写法，其实归并写法，自己并不是很熟练，推介大牛博客

http://blog.163.com/zhaohai_1988/blog/static/20951008520127321239701/

大牛代码真心漂亮!!中间核心，就是学大牛写的

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #define LL long long　　　　　　　　　　　　　　//LL 代替 long long 的写法 中间数据会超出　　int
 using namespace std;

 const int max_size = ;

 int arry[max_size], tmp_arry[max_size];

 LL Merge(int *arr, LL beg, LL mid, LL end, int *tmp_arr)
 {
     memcpy(tmp_arr+beg, arr+beg, sizeof(int)*(end-beg+));
     LL i = beg;
     LL j = mid + ;
     LL k = beg;
     LL inversion = ;
     while(i <= mid && j <= end)
     {
         if(tmp_arr[i] <= tmp_arr[j])   ///如果合并逆序数的时候，前边小于等于后边，就不用记录逆序数的值
         {
             arr[k++] = tmp_arr[i++];
         }else{
             arr[k++] = tmp_arr[j++];   ///如果不是，则要记录逆序数的值
             inversion += (mid - i + );///简单画下就能看出
         }
     }

     while(i <= mid)                    ///把没有并入arr数组的数并入

     {
         arr[k++] = tmp_arr[i++];
     }
     while(j <= end)
     {
         arr[k++] = tmp_arr[j++];
     }
     return inversion;
 }

 LL MergeInversion(int *arr, LL beg, LL end, int *tmp_arr)
 {
     LL inversions = ;
     if(beg < end)
     {
         LL mid = (beg + end) >> ;
         inversions += MergeInversion(arr, beg, mid, tmp_arr);    ///分成两段分别进行记录，递归的进行下去，找逆序数和
         inversions += MergeInversion(arr, mid+, end, tmp_arr);
         inversions += Merge(arr, beg, mid, end, tmp_arr);
     }
     return inversions;
 }

 int main()
 {
     LL n;

     while(cin >> n)
     {
         if(n == )
             break;
         for(int i = ; i < n; ++i)
             scanf("%d", &arry[i]);
         memcpy(tmp_arry, arry, sizeof(int)*n);
         cout << MergeInversion(arry, , n-, tmp_arry) << endl;
     }
     return ;
 }