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 = 500010;

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+1));
    LL i = beg;
    LL j = mid + 1;
    LL k = beg;
    LL inversion = 0;
    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 + 1);///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 = 0;
    if(beg < end)
    {
        LL mid = (beg + end) >> 1;
        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+1, end, tmp_arr);
        inversions += Merge(arr, beg, mid, end, tmp_arr);
    }
    return inversions;
}

int main()
{
    LL n;

    while(cin >> n)
    {
        if(n == 0)
            break;
        for(int i = 0; i < n; ++i)
            scanf("%d", &arry[i]);
        memcpy(tmp_arry, arry, sizeof(int)*n);
        cout << MergeInversion(arry, 0, n-1, tmp_arry) << endl;
    }
    return 0;
}