POJ 2299 Ultra-QuickSort【树状数组】

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536K

Total Submissions: 64049 Accepted: 23919

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

题意：给出一组数据，求其通过快排成有序的操作数。

快速排序的底部操作是选择排序。每进行一次交换，解决一个或多个逆序。

换句话说，整个逆序对的总量，就是答案。

本题有多种解法：（树状数组、线段树、快速排序、归并排序）

1.树状数组：1370ms

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define lowbit(x) (x&(-x))//求最低位数字
const int maxn = 5e5 + 5;
#define LL long long int
struct Data
{
int id, w;

}num[maxn];

int n,  a[maxn];

bool cmp(Data a, Data b)
{
return a.w > b.w;
}

void add(int i)
{
while (i <= n)
{
a[i] += 1;
i += lowbit(i);
}
}
LL sum(int i)
{
LL ans = 0;
while (i > 0)
{
ans += a[i];
i -= lowbit(i);
}
return ans;
}

int main()
{
int i;
while (cin >> n&&n)
{
memset(a, 0, sizeof(a));
for (int i = 0; i < n; i++)
{
num[i].id = i + 1;
cin >> num[i].w;
}
sort(num, num + n, cmp);
LL ans = 0;
for (int i = 0; i < n; i++)
{
ans += sum(num[i].id - 1);//+逆序数
add(num[i].id);
}
cout << ans << endl;
}
return 0;
}