POJ 2299 Ultra-QuickSort

离散化+树状数组求逆序数

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 35024		Accepted: 12608

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 ≤ 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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

struct node
{
    int v,ord;
}s[550000];

bool cmp(node a,node b)
{
    return a.v<b.v;
}

int ar[550000],tree[550000],n;

int lowbit(int x)
{
    return x&-x;
}

void update(int x,int v)
{
    while(x<=n)
    {
        tree[x]+=v;
        x+=lowbit(x);
    }
}

int getsum(int x)
{
    int sum=0;
    while(x>0)
    {
        sum+=tree[x];
        x-=lowbit(x);
    }
    return sum;
}

int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&s.v);
            s.ord=i;
        }
        sort(s+1,s+n+1,cmp);
        for(int i=1;i<=n;i++)
            ar[s.ord]=i;
        memset(tree,0,sizeof(tree));
        long long int ans=0;
        for(int i=1;i<=n;i++)
        {
            update(ar,1);
            ans+=i-getsum(ar);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )