poj 2299 Ultra-QuickSort （树状数组＋离散化）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 48257		Accepted: 17610

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

这道题可以总结的地方不少。

１：对于一组乱序数列，每次只能交换相邻元素，达到有序交换的次数就是原数列中你逆序对的个数。

　　cf上好像总喜欢出这个题。。。我印象中就出现三次了。。。。。

２：原始数组a[i]和树状数组的t[i]的对应问题（？？？存在疑问。。。应该只是这道题，而不是一般规律！）

　　这道题n是５０００００，如果直接开数组是可以开得下的，不需要离散化。但是树状数组的下标对应的是原始数组的值！也就是t[i]的下表最大可能为９９９，９９９，９９９　！　显然存不下，需要离散化。

３：学习了离散化的又一种写法。

/*************************************************************************
    > File Name: code/poj/2299.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月04日 星期二 12时27分32秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=5E5+7;
int n;
int t[N];
int ref[N];

struct Q
{
    int val,id;
}q[N];


bool cmp(Q a,Q b)
{
    if (a.val<b.val) return true;
    return false;
}
int lowbit( int x)
{
    return x&(-x);
}
void update ( int x,int c)
{
    for ( int i = x ; i < N ; i = i + lowbit(i))
    {
    t[i] = t[i] + c;
    }
}
LL Sum( int x)
{
    LL res = 0;
    for ( int i = x; i >= 1 ; i = i - lowbit(i))
    {
    res = res + t[i];
    }
    return res;
}
int main()
{
    while (~scanf("%d",&n)&&n)
    {
    memset(t,0,sizeof(t));
    for ( int i = 1 ; i  <= n ; i++ )
    {
        scanf("%d",&q[i].val);  //离散化的时候相对大小不能打乱
        q[i].id = i;
    }
    sort(q+1,q+n+1,cmp);
    for ( int i = 1 ; i <= n ; i++ )
    {
        ref[q[i].id] = i;
    }
    LL ans = 0;
    for ( int i = 1 ; i <= n ; i++ )
    {
        update(ref[i],1);
        ans = ans + i-Sum(ref[i]);  
      //  cout<<"ans:"<<ans<<endl;
        
    }
    cout<<ans<<endl;
    }
    return 0;
}