# poj2299--B - Ultra-QuickSort（线段树，离散化）

Ultra-QuickSort

 Time Limit: 7000MS Memory Limit: 65536K Total Submissions: 41215 Accepted: 14915

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
```

``````#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL __int64
#define maxn 600000
#define lmin 1
#define rmax n
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define root lmin,rmax,1
#define now l,r,rt
#define int_now LL l,LL r,LL rt
struct node{
LL id1 , id2 ;
LL k ;
}p[maxn] ;
LL cl[maxn<<2] , a[maxn] ;
bool cmp(node a,node b)
{
return a.k < b.k ;
}
void push_up(int_now)
{
cl[rt] = cl[rt<<1] + cl[rt<<1|1] ;
}
void update(LL i,int_now)
{
if( i < l || i > r )
return ;
if( i == l && i==r )
{
cl[rt]++ ;
return ;
}
update(i,lson);
update(i,rson);
push_up(now);
return ;
}
LL query(int ll,int rr,int_now)
{
if( ll > r || rr < l )
return 0;
if( ll <= l && r <= rr )
return cl[rt] ;
return query(ll,rr,lson) + query(ll,rr,rson);
}
int main()
{
LL i , n , m , l , r , x , num ;
while(scanf("%I64d", &m) && m)
{
for(i = 0 ; i < m ; i++)
{
scanf("%I64d", &a[i]);
p[i].k = a[i] ;
p[i].id1 = i ;
}
sort(p,p+m,cmp);
int temp = -1 ;
n = 0 ;
for(i = 0 ; i < m ; i++)
{
if( p[i].k == temp )
p[i].id2 = n ;
else
{
p[i].id2 = ++n ;
temp = p[i].k ;
}
}
for(i = 0 ; i < m ; i++)
a[ p[i].id1 ] = p[i].id2 ;
memset(cl,0,sizeof(cl));
num = 0 ;
for(i = 0 ; i < m ; i++)
{
num += (i - query(1,a[i],root));
update(a[i],root);
}
printf("%I64d\n", num);
}
return 0;
}``````

原文作者：B树
原文地址: https://blog.csdn.net/winddreams/article/details/38443555
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