Hello everybody!

I have used an algorythm that i'm not sure there is or isn't
some subtle stupidity.
It is very simple:

You present the grid by the crossroads and for each crossroad
you find the max number of diagonals(MD) that you can go through from
the start point( crossroad 0,0 ) to the concerned crossroad (of
course
this will be the shortest way to it).
Then you can assume that the paths to all the crossroads which are on
north , east or north-east from this crossroad pass through this point
is shorter by  MD*( 200-100*sqrt(2) ) from the standart(with no
diagonals).
I stop explaining here, `cause my english is awlful, and i may confuse
you.

The code is short so i hope you understand it easily.
So if you can prove that it is wrong or can give me some
test with wa, I`ll be very grateful...

#include <iostream.h>
#include <math.h>

void printks();

class point{ //the crossroads
public:
    int x, y; //the coords
    int dgs;  // the number of diagonals that lead to that
crossroad,
                        //if you take the
shortest path
    point(){ dgs=0; }
} ks[110];

int k, n, m, cinx, ciny, tmpdgs, ind=0, indbr;
double long answer;

int main(){
    cin>>n>>m>>k;

    for (int br=0; br<k; br++){
        cin>>cinx>>ciny;

        tmpdgs=0;
        for ( indbr=0; indbr<ind; indbr++ )
        if ( (cinx-1)>=ks[indbr].x && (ciny-1)>=ks[indbr].y )
                if ( ks[indbr].dgs>tmpdgs ) tmpdgs= ks
[indbr].dgs;

        ks[ind].x=cinx;
        ks[ind].y=ciny;
        ks[ind].dgs=tmpdgs+1;
        ind++;
    }

    tmpdgs=0;
    for ( indbr=0; indbr<ind; indbr++ )
        if ( n>=ks[indbr].x && m>=ks[indbr].y && ks
[indbr].dgs>tmpdgs )
            tmpdgs=ks[indbr].dgs;

    answer= (n+m-2*tmpdgs)*100+sqrt(2)*100*tmpdgs ;

    if ( answer-floorl(answer)<=(long double)(0.49999999999) )
        cout<<(long)(floorl(answer))<<endl;
    else
        cout<<(long)(ceill(answer))<<endl;
    return 0;
}