UVa 12488 - Start Grid

Problem link

Keyword: inversion(a little modified), search

Algorithm:

This is an inversion-counting problem just like UVa 299, but the difference is that when checking if pair(i, j) is inversion or not, we need to decide which element appears first.
That is, if ai appears later than aj in the original list, it's considered an inversion.
In my code, I used STL and pointer comparison to find inversions. 

Code:

#include<cstdio>
#include<algorithm>
using namespace std;
inline bool later( int *seq, int src, int cmp ) { return find( seq, seq+25, src ) > find( seq, seq+25, cmp ); }

int main()
{
    int n, bef[ 25 ], aft[ 25 ];

    while( scanf( "%d", &n ) != EOF )
    {
        int inv = 0;
        for( int i = 0; i < n; ++i )
            scanf( "%d", bef+i );
        for( int i = 0; i < n; ++i )
            scanf( "%d", aft+i );
        for( int i = 0; i < n-1; ++i )
            for( int j = i+1; j < n; ++j )
                inv += later( bef, aft[ i ], aft[ j ] );
        printf( "%d\n", inv );
    }

}