poj 2299 Ultra-QuickSort（树状数组+离散化的题目）据说是简单题，不过还是觉得好难。。。

1、http://poj.org/problem?id=2299

2、题目大意：

一个序列有n个数字，他们是无序的，现在要将这些数字交换顺序，使得他们是从小到大排列的，输出最少交换几次，3、思路分析

如果数字个数很小的话，实际上是可以用冒泡排序的，冒泡交换的步数实际上就是所求，但是数据很大，所以不能这么做，看网上解题报告用树状数组解决，作为第一个树状数组的题目，真心觉得好难，还得继续看，

还有一个问题就是除了n非常大不能用冒泡做，还有就是这n个数字也是非常大的，所以要用到离散化

4、题目

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 37365		Accepted: 13438

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

5、AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 500005
int b
,n,c
;
struct node
{
int id;
int x;
} a
;
int cmp(node a,node b)
{
return a.x<b.x;
}
void update(int i,int x)
{
while(i<=n)
{
c[i]+=x;
i+=i&(-i);
}
}
int sum(int i)
{
int tmp=0;
while(i>0)
{
tmp+=c[i];
i-=i&(-i);
}
return tmp;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i].x);
a[i].id=i;
}
sort(a+1,a+n+1,cmp);
//离散化
a[0].x=-1;
for(int i=1; i<=n; i++)
{
if(a[i].x!=a[i-1].x)
b[a[i].id]=i;
else
b[a[i].id]=b[a[i-1].id];
}
long long ans=0;
for(int i=1; i<=n; i++)
{
//每次更新后判断左边比当前数大的数的个数
update(b[i],1);
ans+=sum(n)-sum(b[i]);
}
printf("%lld\n",ans);
}
return 0;
}