POJ 2299 Ultra-QuickSort（逆序数、分治）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 43803		Accepted: 15974

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

解题报告：根据题意是求交换相邻两位至少需要多少次才能让数组升序。其实就是求逆序数，求逆序数其实和归并排序是一样的，就是在归并的过程中，加入逆序数的统计。例如现在有两个升序数组1 4 9 和 2 6 7 ，那么在归并过程中，插入前一个数组不是逆序，而一旦插入后一个数组中元素，2那么对应的逆序数就加2 。 (因为4和9都还没有插入。)

/********************************/
/*Problem:		POJ 2299	    */
/*User: 	    shinelin	    */
/*Memory: 	    7228K		    */
/*Time: 	    438MS		    */
/*Language: 	C++		        */
/********************************/
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <string>
#include <list>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

#define INF 0x7fffffff
#define LL long long
#define MAXN 500005

LL a[MAXN], com[MAXN];
LL cnt = 0;

void swapTimes(int low, int high)
{
if(low == high - 1)
return ;
if(low == high - 2)
{
if(a[low] > a[low + 1])
{
swap(a[low], a[low + 1]);
cnt ++;
}
return ;
}
int mid = (low + high) >> 1;
swapTimes(low, mid);
swapTimes(mid, high);
int i = low, j = mid;
int k = low;    //三个游标
while (k < high)
{
if(j == high|| i < mid && a[i] <= a[j])
{
com[k++] = a[i++];
}

if(k == high) break;
if(i == mid || j < high && a[i] > a[j])
{
com[k++] = a[j++];
cnt += mid - i;     //归并排序的过程中，统计交换次数。
}
}
for (int i = low; i < high; i ++)
{
a[i] = com[i];
}
return ;
}
int main()
{
int n;
while(~scanf("%d", &n), n)
{
cnt = 0;
for (int i = 0; i < n; i ++)
{
scanf("%I64d", a + i);
}
swapTimes(0, n);
printf("%I64d\n", cnt);
//        for (int i = 0; i < n; i ++)
//        {
//            printf("%d ", a[i]);
//        }
//        printf("\n");
}
return 0;
}