poj 2299 Ultra-QuickSort ：归并排序求逆序数

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 34676		Accepted: 12465

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

题目意图很明确了，就是给你一些数，求逆序数，但是数据量很大，普通的n^2求逆序数的方法铁定超时，所以只能用归并排序求逆序数，合并的时候，我们假设两部分为part1和part2，（part1在前，part2在后）这两部分已经排好序了，那么合并part1和part2的时候，如果part1的top1位置的数大于part2的top2位置的数，那么说明part1后面的那些数也都要比part2的top2位置的数大，所以逆序数就是mid到part1位置的距离

#include<stdio.h>
#include<iostream>
using namespace std;

int array[5000001];
__int64 flag = 0;

void merg(int head, int tail)
{
	int mid = (tail + head) / 2 + 1;
	int * new_array = new int[(tail - head) + 1];
	int top1 = head;
	int top2 = mid;
	int i;
	for(i = 0; top1 < mid  && top2 <= tail ; i++)
	{
		if(array[top1] > array[top2])
		{
			new_array[i] = array[top2];
			top2 ++;
		}
		else
		{
			new_array[i] = array[top1];
			flag += top2 - (mid);
			top1 ++;
		}
	}
	if(top1 == mid && top2 <= tail)
	{
		while(top2 <= tail)
			new_array[i++] = array[top2++];
	}
	else if(top1 != mid && top2 > tail)
	{
		while(top1 < mid)
		{
			new_array[i++] = array[top1++];
			flag += tail - (mid) + 1;
		}
	}
	memcpy(&array[head], new_array, sizeof(int) * (tail - head + 1) );
}
void mergsort(int head, int tail)
{
	if(head >= tail)
		return ;
	mergsort(head, (head + tail) / 2);
	mergsort((head + tail) / 2 + 1, tail);
	merg(head, tail);
}
int main()
{
	int n;
	while(scanf("%d", &n), n != 0)
	{
		int i;
		flag = 0;
		for(i = 0; i < n; i++)
			scanf("%d", &array[i]);

		mergsort(0, n - 1);
		printf("%I64d\n", flag);

	}
	return 0;
}