POJ - 2299 Ultra-QuickSort

 

 

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


题目的意思是求一个数列的逆序数，用树状数组可以直接模拟出来。

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

using namespace std;

const int maxn=500005;
int c[maxn],d[maxn],g[maxn];

void update(int x,int v)
{
    for(;x<=maxn;x+=x&(-x))
        g[x]+=v;
}

int getsum(int x)
{
    int sum=0;
    for(;x>0;x-=x&(-x))
        sum+=g[x];
    return sum;
}

int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));
        memset(g,0,sizeof(g));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&c[i]);
            d[i]=c[i];
        }    
        sort(d,d+n); 
        int size=unique(d,d+n)-d;
        for(int i=0;i<n;i++)
            c[i]=lower_bound(d,d+n,c[i])-d+1;
        /*for(int i=1;i<=n;i++)
            cout<<c[i]<<" ";*/
        long long ans=0;
        for(int i=0;i<n;i++)
        {
            update(c[i],1);
            ans+=i+1-getsum(c[i]);
        }
        printf("%lld\n",ans);
    }
    
    
    return 0;    
}