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.
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 ) ;
}
}