Ultra-QuickSort(树状数组+离散化)

Ultra-QuickSort  POJ 2299

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 50495		Accepted: 18525

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
树状数组维护：c[i]存储比i小的数目.
离散化： 比如输入9 1 0 5 4，将其转化为 5 2 1 4 3，方便存储。
参考别人的代码，看了好半天！

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define LL long long
#define Max 500000+10
using namespace std; 
struct node
{
    int num,order;
};
int n;
int c[Max];        
node in[Max];
int t[Max];
int cmp(node a,node b)
{
    return a.num<b.num;
}
void add(int i,int a)
{
    while(i<=n)
    {
        t[i]+=a;
        i+=i&-i;
    }
}
int sum(int i)
{
    int s=0;
    while(i>=1)
    {
        s+=t[i];
        i-=i&-i;
    }
    return s;
}
int main()
{
    int i,j;
    freopen("in.txt","r",stdin);
    while(~scanf("%d",&n)&&n)
    {
        memset(t,0,sizeof(t));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&in[i].num);
            in[i].order=i;
        }
        sort(in+1,in+1+n,cmp);
        for(i=1;i<=n;i++)             /*离散化*/
            c[in[i].order]=i;
        LL s=0;
        for(i=1;i<=n;i++)
        {
            add(c[i],1);
            s+=i-sum(c[i]);
    //        cout<<s<<endl;
            
        }
        printf("%lld
",s);
    }
}