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

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。

 1 /*************************************************************************
 2     > File Name: code/poj/2299.cpp
 3     > Author: 111qqz
 4     > Email: rkz2013@126.com 
 5     > Created Time: 2015年08月04日 星期二 12时27分32秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<iomanip>
10 #include<cstdio>
11 #include<algorithm>
12 #include<cmath>
13 #include<cstring>
14 #include<string>
15 #include<map>
16 #include<set>
17 #include<queue>
18 #include<vector>
19 #include<stack>
20 #define y0 abc111qqz
21 #define y1 hust111qqz
22 #define yn hez111qqz
23 #define j1 cute111qqz
24 #define tm crazy111qqz
25 #define lr dying111qqz
26 using namespace std;
27 #define REP(i, n) for (int i=0;i<int(n);++i)  
28 typedef long long LL;
29 typedef unsigned long long ULL;
30 const int inf = 0x7fffffff;
31 const int N=5E5+7;
32 int n;
33 int t[N];
34 int ref[N];
35 
36 struct Q
37 {
38     int val,id;
39 }q[N];
40 
41 
42 bool cmp(Q a,Q b)
43 {
44     if (a.val<b.val) return true;
45     return false;
46 }
47 int lowbit( int x)
48 {
49     return x&(-x);
50 }
51 void update ( int x,int c)
52 {
53     for ( int i = x ; i < N ; i = i + lowbit(i))
54     {
55     t[i] = t[i] + c;
56     }
57 }
58 LL Sum( int x)
59 {
60     LL res = 0;
61     for ( int i = x; i >= 1 ; i = i - lowbit(i))
62     {
63     res = res + t[i];
64     }
65     return res;
66 }
67 int main()
68 {
69     while (~scanf("%d",&n)&&n)
70     {
71     memset(t,0,sizeof(t));
72     for ( int i = 1 ; i  <= n ; i++ )
73     {
74         scanf("%d",&q[i].val);  //离散化的时候相对大小不能打乱
75         q[i].id = i;
76     }
77     sort(q+1,q+n+1,cmp);
78     for ( int i = 1 ; i <= n ; i++ )
79     {
80         ref[q[i].id] = i;
81     }
82     LL ans = 0;
83     for ( int i = 1 ; i <= n ; i++ )
84     {
85         update(ref[i],1);
86         ans = ans + i-Sum(ref[i]);  
87       //  cout<<"ans:"<<ans<<endl;
88         
89     }
90     cout<<ans<<endl;
91     }
92     return 0;
93 }