card

#include <fstream>
using namespace std;
int xd[]={-1,-1,0,1,1,1,0,-1},
    yd[]={0,1,1,1,0,-1,-1,-1};
long long val;
int n,m,k,a[52][52],viz[52][52],xstart,ystart,nrsol,sol[20],nr,s[20],nrsolutii;

void copie(int s[],int sol[],int nrsol)
{
    for(int i=1;i<=nrsol;++i)
        sol[i]=s[i];
}

void bordare()
{
    int i;
    for(i=0;i<=n+1;++i)
        viz[i][0]=viz[i][m+1]=-1;
     for(i=0;i<=m+1;++i)
        viz[0][i]=viz[n+1][i]=-1;
}

void citire()
{
    ifstream f("card.in");
    int i,j;
    f>>n>>m>>xstart>>ystart>>k;
    for(i=1;i<=n;++i)
        for(j=1;j<=m;++j)
            f>>a[i][j];
    f.close();
}

void becktracking(int x,int y)
{
    int xn,yn;
    if(val==a[xstart][ystart])
    {
        if(nr<nrsol||nrsol==k+1)
        {
            nrsol=nr;nrsolutii=1;k=nr;
            copie(s,sol,nrsol);
        }
        else
            if(nr==nrsol) /// solutia s are nr elemente; sol optima sol are nrsol elemente
            {
                ++nrsolutii; /// nrsolutii retine numar de solutii obtinute
                if(s[nr]<sol[nrsol])
                    copie(s,sol,nrsol);
                else
                    if(s[nr]==sol[nrsol]&&s[1]<sol[1])
                        copie(s,sol,nrsol);
            }
    }
    else
        for(int dir=0;dir<8;dir++)
        {
            xn=x+xd[dir];yn=y+yd[dir];
            if(viz[xn][yn]==0&&nr<k)
            {
                viz[xn][yn]=1;
                for(int i=0;i<4;i++) /// aleg operatia
                {
                    switch(i)
                    {
                        case 0:val+=a[xn][yn];break;
                        case 1:val-=a[xn][yn];break;
                        case 2:val+=2*a[xn][yn];break;
                        case 3:val+=(a[xn][yn]/2);break;
                    }
                    ++nr;
                    s[nr]=a[xn][yn];
                    becktracking(xn,yn);
                    --nr;
                    switch(i)
                    {   case 0:val-=a[xn][yn];break;
                        case 1:val+=a[xn][yn];break;
                        case 2:val-=2*a[xn][yn];break;
                        case 3:val-=(a[xn][yn]/2);break;
                    }
                }
                viz[xn][yn]=0;
            }
       }
}

int main()
{
    int i;
    citire();
    bordare();
    viz[xstart][ystart]=1;
    nr=0;
    nrsol=k+1;
    becktracking(xstart,ystart);
    ofstream g("card.out");
    g<<nrsolutii<<'\n';
    for(i=1;i<=nrsol;++i)
        g<<sol[i]<<' ';
    g.close();
    return 0;
}