poj 2299 Ultra-QuickSort（树状数组求逆序数+离散化）

题目链接：http://poj.org/problem?id=2299

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

逆序数。

代码例如以下：

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn=500017; int n; int aa[maxn]; //离散化后的数组 int c[maxn]; //树状数组struct Node {int v;int order; }in[maxn];int Lowbit(int x) //2^k {return x&(-x); }void update(int i, int x)//i点增量为x {while(i <= n){c[i] += x;i += Lowbit(i);} } int sum(int x)//区间求和 [1,x] {int sum=0;while(x>0){sum+=c[x];x-=Lowbit(x);}return sum; }bool cmp(Node a ,Node b) {return a.v < b.v; }int main() {int i,j;while(scanf("%d",&n) && n){//离散化for(i = 1; i <= n; i++){scanf("%d",&in[i].v);in[i].order=i;}sort(in+1,in+n+1,cmp);for(i = 1; i <= n; i++) aa[in[i].order] = i;//树状数组求逆序memset(c,0,sizeof(c));__int64 ans=0;for(i = 1; i <= n; i++){update(aa[i],1);ans += i-sum(aa[i]);//逆序数个数}printf("%I64d\n",ans);}return 0; }

转载于:https://www.cnblogs.com/zsychanpin/p/6919683.html

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