POJ 2299 Ultra-QuickSort

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

/*
      去年寒假集训的最后一周周赛题目，当时没做出来，知道是求将序列转化为有序序列需要交换的次数，当时想到了冒泡和逆序数，一直过不了，赛后学长讲题的时候说可以用树状数组加离散化处理，也可以用归并排序，也就是求序列的逆序数，7000MS明显坑人呀..
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stack>
#include <cmath>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;

struct node
{
    int x;
    int pos;
}a[500100];
int L[500100],C[500100];
long long ans;
bool cmp(node x,node y)
{
    return x.x < y.x;
}
void up(int x,int n)
{
    while(x <= n)
    {
        C[x]+=1;
        x+=(x&(-x));
    }
}
int get(int x)
{
    int sum = 0;
    while(x > 0)
    {
        sum+=C[x];
        x-=(x&(-x));
    }
    return sum;
}
int main()
{
    int n;
    while(~scanf("%d",&n) && n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].x);
            a[i].pos = i;
        }
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++)
            L[a[i].pos] = i;
        ans = 0;
        memset(C,0,sizeof(C));
        for(int i=1;i<=n;i++)
        {
            up(L[i],n);
            ans+=i-get(L[i]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}