poj 2299 Ultra-QuickSort (树状数组)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 37971		Accepted: 13672

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

//Accepted    9356K    563MS    C++    1145B
/*

    题意：
        给出一串数列，问用冒泡排序发要进行多少次交换。
    
    树状数组：  
        我的理解是这题和poj 2352类似，不过这题一定要加上离散化。
       
        这题就是要求每个元素中排在其前面且比它大的元素的数量的和。
        我转换成和之前那题一样的思想，由最大的数减去该数的排位，
    最后求出来的a[i]就是所求，表示前面有i个比它大的数。然后每个
    数都要交换i次，故要乘上i。 
     

*/
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#define N 500005
using namespace std;
struct node{
   int val;
   int pos;
};
node no[N];
int b[N],a[N],c[N];
int cmp(const void*a,const void*b)
{
    return (*(node*)a).val-(*(node*)b).val;
}
inline int lowbit(int i)
{
    return i&(-i);
}
void update(int k,int detal)
{
    for(;k<N;k+=lowbit(k)) 
        c[k]+=detal;
}
inline int getsum(int k)
{
    int t=0;
    for(;k>0;k-=lowbit(k)) 
        t+=c[k];
    return t; 
}
int main(void)
{
    int n;
    while(scanf("%d",&n)!=EOF && n)
    {
        __int64 sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&no[i].val);
            no[i].pos=i;
        }
        qsort(no,n,sizeof(no[0]),cmp);
        for(int i=0;i<n;i++) b[no[i].pos]=i;
        //for(int i=0;i<n;i++) printf("%d
",b[i]);
        memset(c,0,sizeof(c));
        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++){
            int t=N-b[i]-5;
            a[getsum(t)]++;
            update(t,1);
        }
        for(int i=1;i<n;i++) sum+=i*a[i];
        printf("%I64d
",sum);
    }
    return 0;
}