Poj 2299 - Ultra-QuickSort 离散化,树状数组,逆序对 - Html/Css

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 52306		Accepted: 19194

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：给定n个数，只能交换相邻的两个元素，至少交换几次，成为递增序列。

题解：

明显要求序列的逆序对数目。。。

对于样例：

9 1 0 5 4

我们将其排序：

0 1 4 5 9

在每个位置上初始放为1.

1 1 1 1 1

然后，从原序列开始遍历。

先到9，我们把其排好序的位置拿出，即为5。

然后统计位置5之前有多少1。

1 1 1 1 1

———— > 4个 ans+=4

然后把5号位置放为0。

1 1 1 1 0

继续操作即可。。。

这个树状数组维护即可。。。

注意开long long和原序列排序后要去重。。。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 500010
#define LL long long
int n,BIT[MAXN],a[MAXN],aa[MAXN];
int read()
{
int s=,fh=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}
while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
return s*fh;
}
int Lowbit(int o){return o&(-o);}
void Update(int o,int o1)
{
while(o<=n)
{
BIT[o]+=o1;
o+=Lowbit(o);
}
}
int Sum(int o)
{
int sum=;
while(o>)
{
sum+=BIT[o];
o-=Lowbit(o);
}
return sum;
}
int main()
{
int tot,i,wz;
LL ans;
while()
{
n=read();if(n==)break;
for(i=;i<=n;i++){a[i]=read();aa[i]=a[i];}
memset(BIT,,sizeof(BIT));
sort(a+,a+n+);
tot=unique(a+,a+n+)-(a+);
for(i=;i<=tot;i++)Update(i,);
ans=;
for(i=;i<=n;i++)
{
wz=lower_bound(a+,a+tot+,aa[i])-a;
ans+=(LL)Sum(wz-);
if(BIT[wz]!=)Update(wz,-);
}
printf("%lld\n",ans);
}
fclose(stdin);
fclose(stdout);
return ;
}