poj 2299 Ultra-QuickSort（樹狀數組求逆序數+離散化）

標籤：逆序數   離散化   

題目連結：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

59105431230

Sample Output

60

Source

Waterloo local 2005.02.05

逆序數；

代碼如下：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn=500017;int n;int aa[maxn];   //離散化後的數組int c[maxn];    //樹狀數組 struct Node{   int v;   int order;}in[maxn]; int Lowbit(int x) //2^k{    return x&(-x);}void update(int i, int x)//i點增量為x{while(i <= n){c[i] += x;i += Lowbit(i);}}int sum(int x)//區間求和 [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)    {        //離散化        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;        //樹狀數組求逆序        memset(c,0,sizeof(c));        __int64 ans=0;        for(i = 1; i <= n; i++)        {            update(aa[i],1);            ans += i-sum(aa[i]);//逆序數個數        }printf("%I64d\n",ans);    }    return 0;}