poj 2299 Ultra-QuickSort ：归并排序求逆序数

点击打开链接

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;
}