HDU 2838 cow sorting 树状数组

       话说这道题要用的三个树状数组，不容易啊。我刚开始想的时候想明白了用公式怎么算，却想不出来怎么转化到树状数组上，总感觉有些地方实现不了，原来竟然是用三个树状数组。。。这让只写过一个树状数组的孩纸情何以堪？

          具体来说，有一个num数组，里面记录的是插入a[i]后，在a[i]之前插入且比a[i]小的数的个数；还有一个totalsum数组，记录的是插入第i个数后，前i-1个数的总和；还有一个smallersum数组，记录的是插入a[i]后，在a[i]之前插入且比a[i] 小的数的总和。这样最后就可以算出来了。题目：

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1173    Accepted Submission(s): 385

Problem Description

Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage Sherlock's milking equipment, Sherlock would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes Sherlock a total of X + Y units of time to exchange two cows whose grumpiness levels are X and Y.

Please help Sherlock calculate the minimal time required to reorder the cows.

 

Input

Line 1: A single integer: N
Lines 2..N + 1: Each line contains a single integer: line i + 1 describes the grumpiness of cow i.

 

Output

Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.

 

Sample Input

   
   
   
   

    
    
    
    3
2
3
1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7


    
    
    
    
 
     
 
      Hint 
     
 Input Details Three cows are standing in line with respective grumpiness levels 2, 3, and 1. Output Details 2 3 1 : Initial order. 2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4). 1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3). 
    
    
    
    
     
   
   
   
   

#include <cstdio>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int M = 100010;
__int64 num[M] , totalsum[M] , smallsum[M];

__int64 lowbit( __int64 x )
{
        return x&(-x);        
}

void add( __int64 x[] , __int64 xx , __int64 yy )
{
     while( xx < M )
     {
            x[xx] += yy;
            xx +=lowbit(xx);
     }
}

__int64 sum( __int64 x[] , __int64 xx )
{
        __int64 ss = 0;
        while( xx > 0 )        
        {
               ss += x[xx];
               xx -= lowbit(xx);       
        }
        return ss;
}

int main( )
{
    __int64 n;
    while( ~scanf("%I64d",&n) )
    {
           memset( num , 0 , sizeof(num) );
           memset( totalsum , 0 , sizeof(totalsum) );
           memset( smallsum , 0 , sizeof(smallsum) );
           __int64 x , ans = 0 , count = 0;
           for( __int64 i=1 ; i<=n ; i++ ){
                scanf("%I64d",&x);
                add( num , x , 1 );
                add( totalsum , i , x );
                add( smallsum , x , x );
                count = sum( num , x-1 );
                ans +=  (i-count-1)*x;
                ans += sum(totalsum,i-1);
                ans -= sum(smallsum,x-1);
           }
           printf("%I64d\n",ans);
    }
    return 0;
}