poj2299(逆序数+线段树)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 34679		Accepted: 12466

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

 

        本题给定一个序列，要求相邻的数交换多少次后可以得到一个递增的序列。这个题的过程似曾相识，仔细想了想，原来是冒泡排序的基本过程思想。冒泡模拟，肯定会超时的，冒泡的时间复杂度为O(N*N)。我们可以这样想，我们先把最大的移到最后，然后是次大的……这样一来就变了了我们熟悉的模型，求该序列的逆序数了。求逆序数是经典的线段树的用法。

         这里说说求逆序数的线段树的做法。先把序列离散化，即从小到大排序，序号与数列中的数对应，然后按序列原有的顺序依次统计每个数的对应的逆序数，并修改线段树。具体做法是先查询线段数中比它大且排在它前面的数，对应区间为当前到末尾的区间和；然后再对应位置插入该数，即将对应位置修改为1.下面使用树状数组实现的：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=500000+100;
int da[MAXN];
int tda[MAXN];
//**********************************************
int C[MAXN];
int Lowbit[MAXN];
//C[i] = a[i-lowbit(i)+1] + …+ a[i],下表从1开始
//Lowbit[i]=i&(i^(i-1));或Lowbit[i]=i&(-i); 
//1.查询
int QuerySum(int p)
//查询原数组中下标1-p的元素的和 
{ 
   int nSum=0; 
   while(p>0) 
   {  
      nSum+=C[p]; 
      p-=Lowbit[p]; 
   } 
    return nSum; 
} 

//2.修改+初始化
void Modify(int p,int val) 
//原数组中下表为p的元素+val，导致C[]数组中部分元素值的改变
{ 
    while(p<=MAXN-10) 
   { 
      C[p]+=val; 
      p+=Lowbit[p]; 
    } 
} 
//************************************************************

bool cmp(int a,int b)
{
	return a<b;
}

int Binary_research(int tmp,int n)
{
	int mid,low=1,high=n;
	while(low<=high)
	{
		mid=(low+high)>>1;
		if(tda[mid]<=tmp)low=mid+1;
		else high=mid-1;
	}
	return high;
}

int main()
{
	int n,i;
	__int64 ans;
	for(i=1;i<MAXN;i++)
		Lowbit[i]=i&(-i);

	while(~scanf("%d",&n),n)
	{
		for(i=1;i<=n;i++)
		{
			scanf("%d",&da[i]);
			tda[i]=da[i];
		}
		sort(tda+1,tda+n+1,cmp);
		memset(C,0,sizeof(C));

		ans=0;
		for(i=1;i<=n;i++)
		{
			int id=Binary_research(da[i],n);
			ans+=(__int64)(QuerySum(n)-QuerySum(id));
			Modify(id,1);
		}
		printf("%I64d\n",ans);
	}
	return 0;
}