hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6815    Accepted Submission(s): 4158

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

 

Output

For each case, output the minimum inversion number on a single line.

 

 

Sample Input

10 1 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16

代码 +　 注释　

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#define maxn 5001
using namespace std ;
int xt[maxn] , a[maxn];
int low( int x )
{
	return x & (-x) ;
}
void up( int x )
{
	while( x <= maxn )
    { 
		xt[x] += 1 ;
		x += low(x) ;
	}
}
int sum( int x )
{
	int ans = 0 ;
	while( x > 0 )
	{
		ans += xt[x] ;
		x -= low(x) ;
	}
	return ans ;
}
int main()
{
	int i , j , n ,ans , T , m = 0 ;
	while(scanf( "%d" , &n ) !=EOF )
	{
		memset( xt , 0 ,sizeof(xt) ) ;
		ans = 0 ;
		for( i = 1 ; i <= n ;i++ )
		{
			scanf( "%d" , &a[i] ) ;
			a[i]+=2 ;
			up(a[i]) ;
			ans += i - sum(a[i]-1) - 1 ;
		}
		m = ans ;
		for( i = 1 ;i < n ;i++ ){
			a[i] -= 2 ;
			// 下面是规律
	    	 m = m + (n - 1 - a[i]) - a[i];  
            if( m < ans)  
                ans = m ;  			
		}
		printf( "%d\n" , ans ) ;
	}
}