208 lines
7.3 KiB
C
208 lines
7.3 KiB
C
#include <stdio.h>
|
|
#include <string.h>
|
|
#include <assert.h>
|
|
#include <stdlib.h>
|
|
#include <stdbool.h>
|
|
#define SIZE 100
|
|
|
|
// Struktura na vytvorenie bynarneho stromu. Ma ulozenu hodnotu, pole charov, a referencii na l'aveho a praveho syna.
|
|
// V danom pripade je to plny bynarny strom, cize kazdy uzol bude mat dva alebo nullu synov
|
|
struct tree{
|
|
char value[SIZE];
|
|
/// Referencii v pamati
|
|
struct tree* left;
|
|
struct tree* right;
|
|
};
|
|
|
|
// Funkcia na uvolnenie bynarneho stromu. Je to recursivna implementacia, znamena, ze danu funkciu budeme volat vo vnuri seba na iny argument.
|
|
// Tato funkcia a vsetky ine recursivne funkcii v kontexte binarneho stromu operuju na baze postorder, cize budeme navstevat' na zaciatku podstromy a potom sam uzol
|
|
// Pre-order operuje tak, ze na zaciatku ide sam uzol a potom podstormy, a inorder - jeden podstrom, uzol iny podstrom
|
|
void destroy_tree(struct tree* tree){
|
|
/// V danom pripade, uvolnime l'avy podstrom, potom pravy podstrom a koren
|
|
if(tree->left){
|
|
// dojde do uzla bez synov, ho uvolnime a uvolnime vsetky ine uzly v podstrome
|
|
destroy_tree(tree->left);
|
|
}
|
|
if(tree->right){
|
|
// Tak tiez, prechadzame do uzla bez potomkov pravom podstrome
|
|
destroy_tree(tree->right);
|
|
}
|
|
//Vlasne uvolnime v pamati uzol
|
|
free(tree);
|
|
}
|
|
// Funkcia pre nacitanie stromu
|
|
struct tree* read_tree(){
|
|
///Buffer pre nacitanie
|
|
char buffer[SIZE];
|
|
/// Inticializacia vsetkych bytov na 0
|
|
memset(buffer,0,SIZE);
|
|
/// vlasne nacitanie jedneho riadku. Ako argumenty mame kam, pocet bytov a odkial'
|
|
char* r = fgets(buffer,SIZE,stdin);
|
|
// Vratime NULL ak nepodarilo nacitat'
|
|
if(!r) {
|
|
return NULL;
|
|
}
|
|
/// Odmietnime newline a premenime ho na null terminator - koniec riadku
|
|
buffer[strcspn(buffer, "\n")] = '\0';
|
|
/// Allokujeme v pamati jeden uzol. Calloc inticializuje vsetky byty na 0.
|
|
struct tree* node = calloc(1,sizeof(struct tree));
|
|
///Prekopirujeme naticany riadok v pamati do hodnoty uzla
|
|
memcpy(node->value,buffer,SIZE);
|
|
// Ak ma riadok *, tak to znamena ze to odpoved - listovy uzol a nebude mat' synov
|
|
if(buffer[0] == '*'){
|
|
// Syny su NULL
|
|
node->left = NULL;
|
|
node->right = NULL;
|
|
// vratime uzol
|
|
return node;
|
|
|
|
}
|
|
// Pokracujeme v nacitani, ak nie je to odpoved
|
|
node->left = read_tree();
|
|
node->right = read_tree();
|
|
/// Ak sconcime nacitanie, tak vratime koren
|
|
return node;
|
|
}
|
|
/// Pomocna funkcia na vyhl'adovanie nepravidelneho vstupu
|
|
bool find_the_incorrect(struct tree* tree) {
|
|
if (tree == NULL)
|
|
return false;
|
|
|
|
if /// Ak strcspn, funkcia ze najdi kde prvy krat tento pattern, vrati index, tak to znamena, ze tieto chary su v slove
|
|
// Ak symboly nie su v ret'azce, tak to vratilo by dlzku retazca
|
|
// Overenie pre nekoretne otazky: nemozu mat' * na zacatku
|
|
((strcspn(tree->value, "*") != strlen(tree->value) &&
|
|
strcspn(tree->value, "?") != strlen(tree->value)) ) {
|
|
// vratime true ak to je nekorektna otakzka
|
|
return true;
|
|
}
|
|
// prechadzame po vsetkemu stromu a vyhladavame recursivne prvky
|
|
// || - logicky operator OR - alebo
|
|
// Na zacatku pozrieme do l'aveho podstromu a potom do praveho
|
|
return find_the_incorrect(tree->left) || find_the_incorrect(tree->right);
|
|
}
|
|
void print_tree(struct tree* tree,int offset){
|
|
for (int i = 0; i < offset; i++){
|
|
printf(" ");
|
|
}
|
|
printf("%s",tree->value);
|
|
if (tree->left){
|
|
print_tree(tree->left,offset +3);
|
|
print_tree(tree->right,offset +3);
|
|
}
|
|
}
|
|
|
|
// Funkcia pre pocitanie listov binarneho stromu. Recursivna implementacia.
|
|
// List - je to uzol bez synov. Musi ich byt na 1 viac ako vnutornych uzlov, tych, co maju potomkov.
|
|
int count_leaves(struct tree* tree){
|
|
// Na zacatku pocet je null
|
|
int count = 0;
|
|
if(tree == NULL){
|
|
// Ak koren je NULL, tak to znamena ze strom je NULL a vratime pocet, cize 0
|
|
return count;
|
|
}
|
|
// Inak, ak je len jeden koren - vratime 1
|
|
if(tree->left == NULL && tree->right == NULL && tree){
|
|
return count +1;
|
|
}
|
|
// Inak prechadzame strom, na zaciatku volame funkciu na l'avy podstom a potom a pravy
|
|
else{
|
|
return count + count_leaves(tree->left) + count_leaves(tree->right);
|
|
|
|
}
|
|
}
|
|
|
|
/// Tak tiez, recursivna funkcia na pocet ne lisov
|
|
int count_non_leaves(struct tree* tree){
|
|
// Ak je uzol NULL alebo ne ma synov, tak vratime 0
|
|
if(tree == NULL || ( tree->left== NULL && tree->right ==NULL)){
|
|
return 0;
|
|
}
|
|
// Inak, vratime 1(sam uzol) + pocet ne uzlov v oboch podstromoch
|
|
return 1 + count_non_leaves(tree->left) +count_non_leaves(tree->right);
|
|
}
|
|
|
|
|
|
/// Main pre systemem
|
|
int main(void){
|
|
/// Overenie na medzeru po databaze
|
|
bool whitespace = false;
|
|
/// Vytvorime strom
|
|
struct tree* tree = read_tree();
|
|
/// Buffer pre prazdny riadok
|
|
char line[SIZE];
|
|
/// Spocitame listy a vnutorny uzly
|
|
int numOfLeaves = count_leaves(tree);
|
|
int numOfNonLeaves = count_non_leaves(tree);
|
|
/// Pole pre odpoved' od pouzivatel'a
|
|
char answer[SIZE];
|
|
// Hl'adame prazdny riadok pomcou while
|
|
while (fgets(line, sizeof(line), stdin)) {
|
|
// Pokial' mame co nacitovat' po databazu pre strom
|
|
/// Musime odmietnut newline aby spravne porovnat'
|
|
line[strcspn(line, "\n")] = '\0';
|
|
/// Ak prvy prvok bufferu bude null terminator, tak to bude prazdny riadok
|
|
if (line[0] == '\0') {
|
|
// Check je splneny
|
|
whitespace = true;
|
|
// Vyjdeme z cyklusu
|
|
break;
|
|
}
|
|
}
|
|
/// Prechadzame podstromy a hl'adame nespravne vstupy
|
|
bool found = find_the_incorrect(tree);
|
|
printf("Expert z bufetu to vie.\n");
|
|
/// Ak nepodarilo nacitst alebo nie je tam medzera alebo nasli sme nespravny vstup, vypiseme hlasku a sconcime program.
|
|
// To plati ak zaciatok bazy nie je spravny
|
|
/// Ak stringy su rovnake, atak strcmp vrati 0
|
|
if(!tree || !whitespace || found || strcmp(tree->value, "Rastie to nad zemou alebo pod zemou ?") == 0){
|
|
printf("Chybna databaza\n");
|
|
return 0;
|
|
}
|
|
|
|
printf("Pozna %d druhov ovocia a zeleniny.\n", numOfLeaves);
|
|
printf("Odpovedajte 'a' pre prvu moznost alebo 'n' pre druhu moznost.");
|
|
/// Pokial uzol nie je null, pokracuejme v systeme
|
|
while(tree){
|
|
/// Vypis otazky
|
|
printf("\n%s", tree->value);
|
|
/// Nacitame vstup od pouzivatel'a
|
|
fgets(answer, sizeof(answer), stdin);
|
|
// Odmietnime newline pre porovnanie
|
|
answer[strcspn(answer, "\n")] = '\0';
|
|
// Ak je to prazdny ridok, tak to znamena ze vstup uz ne mame
|
|
if(answer[0] == '\0'){
|
|
printf("\nKoniec vstupu\n");
|
|
return 0;
|
|
}
|
|
/// Ak je to nie, tak prejdeme na pravy podstrom
|
|
if(strcmp(answer, "n") == 0){
|
|
tree = tree->right;
|
|
/// Ak hodnotra daneho uzla je odpoved, ju vypiseme a skoncime prgram
|
|
if(tree->value[0] == '*'){
|
|
printf("\n%s\n", tree->value);
|
|
printf("Koniec\n");
|
|
return 0;
|
|
}
|
|
|
|
}
|
|
// Analogicky pre ano
|
|
else if(strcmp(answer, "a") == 0){
|
|
tree = tree->left;
|
|
if(tree->value[0] == '*'){
|
|
printf("\n%s\n", tree->value);
|
|
printf("Koniec\n");
|
|
|
|
return 0;
|
|
}
|
|
|
|
}
|
|
/// Inak, fakticke skoncime program
|
|
else{
|
|
printf("\nNerozumiem\n");
|
|
return 0;
|
|
}
|
|
}
|
|
/// Uvolnime v pamati strom'""
|
|
destroy_tree(tree);
|
|
return 0;} |