poj 2299 Ultra-QuickSort（树状数组求逆序数+离散化）

题目链接：http://poj.org/problem?id=2299

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

逆序数。

代码例如以下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=500017;
int n;
int aa[maxn];   //离散化后的数组
int c[maxn];    //树状数组
 
struct Node
{
   int v;
   int order;
}in[maxn];
 
int Lowbit(int x) //2^k
{
    return x&(-x);
}

void update(int i, int x)//i点增量为x
{
	while(i <= n)
	{
		c[i] += x;
		i += Lowbit(i);
	}
}
int sum(int x)//区间求和 [1,x]
{
	int sum=0;
	while(x>0)
	{
		sum+=c[x];
		x-=Lowbit(x);
	}
	return sum;
}

bool cmp(Node a ,Node b)
{
    return a.v < b.v;
}

int main()
{
    int i,j;
    while(scanf("%d",&n) && n)
    {
        //离散化
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&in[i].v);
            in[i].order=i;
        }
        sort(in+1,in+n+1,cmp);
        for(i = 1; i <= n; i++) 
			aa[in[i].order] = i;
        //树状数组求逆序
        memset(c,0,sizeof(c));
        __int64 ans=0;
        for(i = 1; i <= n; i++)
        {
            update(aa[i],1);
            ans += i-sum(aa[i]);//逆序数个数
        }
		printf("%I64d
",ans);
    }
    return 0;
}