
#include "stdafx.h"

#include "form.h"
#include "tree.h"

#include 
#include 
using namespace std;

const int Max = 4;

struct Item
{
    int cnt; // pocet klicu
    int key [Max+1]; // key[0], ... key[cnt-1], a jeden navic
    Item * ref [Max+2]; // ptr [0], ... ptr [cnt], a jeden navic
    Item();
};

Item::Item()
{
    cnt = 0;
    for (int i = 0; i < Max + 1; i++) key[i] = -1;
    for (int i = 0; i < Max + 2; i++) ref [i] = nullptr;
}

void enter0 (Item * p, int val)
{
    int i = 0;
    while (i < p->cnt && p->key[i] < val) i++;
    // i == p->cnt , vsechny byly mensi, pridat 
    // i < p->cnt, p->key[i] >= val
    if (i < p->cnt && p->key[i] == val) {  }
    else {
        Item * t = p->ref[i];
        if (t == nullptr)
        {
            for (int k = p->cnt - 1; k >= i; k--) p->key[k + 1] = p->key[k];
            for (int k = p->cnt; k > i; k--) p->ref[k + 1] = p->ref[k];
            p->key[i] = val;
            p->ref[i] = nullptr;
            p->cnt++;
        }
        else
        {
            enter0 (t, val);
            if (t->cnt == Max + 1)
            {
                int a = Max / 2; // a klicu zustane v t
                int b = t->cnt - a -1; // b klicu do noveho bloku
                int d = t->key[a]; // delici bod

                Item * n = new Item;
                n->cnt = b;
                for (int k = a + 1; k < t->cnt; k++) n->key[k - a - 1] = t->key[k];
                for (int k = a + 1; k < t->cnt+1; k++) n->ref[k - a - 1] = t->ref[k];

                for (int k = a + 1; k < t->cnt; k++) t->key[k] = -1;
                for (int k = a + 1; k < t->cnt + 1; k++) t->ref[k] = nullptr;
                t->cnt = a;

                for (int k = p->cnt - 1; k >= i; k--) p->key[k + 1] = p->key[k];
                for (int k = p->cnt; k > i; k--) p->ref[k + 1] = p->ref[k];
                p->key[i] = d;
                p->ref[i+1] = n;
                p->cnt++;
            }
        }
    }
}

void enter (Item * & p, int val)
{
    if (p == nullptr)
    {
        p = new Item;
        p->cnt = 1;
        p->key[0] = val;
        p->ref[0] = nullptr;
        p->ref[1] = nullptr;
    }
    else
    {
        enter0 (p, val);
        Item * t = p;
        if (t->cnt == Max + 1)
        {
            int a = Max / 2; // a klicu zustane v t
            int b = t->cnt - a - 1; // b klicu do noveho bloku
            int d = t->key[a]; // delici bod

            Item * n = new Item;
            n->cnt = b;
            for (int k = a + 1; k < t->cnt; k++) n->key[k - a - 1] = t->key[k];
            for (int k = a + 1; k < t->cnt + 1; k++) n->ref[k - a - 1] = t->ref[k];

            for (int k = a + 1; k < t->cnt; k++) t->key[k] = -1;
            for (int k = a + 1; k < t->cnt + 1; k++) t->ref[k] = nullptr;
            t->cnt = a;

            p = new Item;
            p->cnt = 1;
            p->ref[0] = t;
            p->key[0] = d;
            p->ref[1] = n;
        }
    }
}
void display (Item * p)
{
    if (p != nullptr) 
    {
        openBranch ("branch");
        for (int i = 0; i < p->cnt; i++)
        {
            display (p->ref[i]);
            addNode(to_string(p->key[i]));
        }
        display(p->ref[p->cnt]);
        closeBranch();
    }
}

void run()
{
    Item * root = nullptr;
    enter(root, 30);
    enter(root, 20);
    enter(root, 10);
    enter(root, 40);
    enter(root, 50);

    for (int k = 21; k <= 29; k++) enter(root, k);

    for (int k = 51; k <= 800; k++) enter(root, k);

    display(root);

    status ("O.K.");
}