poj 2299 Ultra-QuickSort(BIT求逆序数)

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
解题思路:题意很简单,树状数组or归并排序求逆序数,这里只讲树状数组的实现!因为a[i]的值高达10^9,树状数组的大小肯定开不了这么大,而n最大为5e5(可作为数组大小,不过大小还要再开大一点,避免越界),因此需要将原来每个元素离散化,即重新编个号(1~n)。
做法:用一个结构体记录原来每个元素val出现的次序id,然后将结构体按val的大小升序排序,接下来遍历排序后的结构体数组,将原来的元素离散化成1~n,即将id对应原来的数字改成第i大(i∈[1,n]),最后就可以直接用树状数组进行更新和统计逆序数了。
拿题目中9 1 0 5 4这个样例来加强对数据离散化的理解:
输入的元素值 9 1 0 5 4 -->排序后 0 1 4 5 9
对应的次序id 1 2 3 4 5 3 2 5 4 1
此时将排序后每个id对应的元素离散化成第i小即 1 2 3 4 5,显然0是第1小,且是第3次出现,1是第2小,且是第2次出现...
这样我们就已经成功地把原来的数据离散化,接下来遍历一下次序id:tar[1]=5(原来为9,9是第一个输入的,这里就变成了5,空间上压缩了不少),先在树状数组中标记为1,并且5前面有4个空为0,于是5(9)这个元素构成了4个逆序对,累加逆序数4并继续按此操作下去即可找出所有的逆序数。
AC代码:
 1 #include<iostream>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn=500005;
 6 typedef long long LL;
 7 int n,val,aa[maxn],tar[maxn];
 8 struct node{int val,id;}nod[maxn];
 9 bool cmp(node a,node b){return a.val<b.val;}
10 int lowbit(int x){
11     return x & -x;
12 }
13 void update(int x,int val){
14     while(x<=n){
15         aa[x]+=val;
16         x+=lowbit(x);
17     }
18 }
19 int getsum(int x){
20     int ret=0;
21     while(x>0){
22         ret+=aa[x];
23         x-=lowbit(x);
24     }
25     return ret;
26 }
27 int main(){
28     while(cin>>n&&n){
29         LL ans=0;
30         memset(aa,0,sizeof(aa));//注意清空
31         for(int i=1;i<=n;++i){
32             cin>>nod[i].val;
33             nod[i].id=i;//记录元素val出现的次序id
34         }
35         sort(nod+1,nod+n+1,cmp);//然后数组元素val按升序排序
36         for(int i=1;i<=n;++i)tar[nod[i].id]=i;//离散化数据:tar[nod[i].id]表示原来第nod[i].id次出现的值换成现在1~n中的编号i
37         for(int i=1;i<=n;++i){
38             update(tar[i],1);//tar[i]表示为输入值的次序:第i次出现的值(已离散化),先将该值在树状数组中标记为1,表示该数字已出现
39             ans+=tar[i]-getsum(tar[i]);//求出tar[i]前面还没出现数字的个数即为与当前tar[i]构成逆序对的个数,然后累加即可
40         }
41         cout<<ans<<endl;
42     }
43     return 0;
44 }

 

posted @ 2018-08-14 18:35  霜雪千年  阅读(283)  评论(0编辑  收藏  举报