50240. Cubic Tower and String
難度：5/5
這種tower類型的題目，最重要的是要怎麼設計你的3d array，讓你找到你這一次要消掉的東西的時候，可以用\(O(1)\)找到你要insert下一個的地方。1
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111#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define max(a,b) ((a)>(b)?(a):(b))
#define left(n) (2 * (n))
#define right(n) (2 * (n) + 1)
#define parent(n) ((n) / 2)
int n, m;
char tower[102 * 102 * 102] = {'\0'};
int swap(int *a, int *b){
int t = *a;
*a = *b;
*b = t;
}
int get_id(int x, int y, int z){
return x * (102 * 102) + y * (102) + z;
}
static inline int cmp_str_id(int id1, int id2){
return strcmp(&(tower[id1]), &(tower[id2]));
}
typedef struct{
int a[10050];
int size;
} heap;
int get_largest(heap* h, int i){
int l = left(i);
int r = right(i);
int largest;
if(l <= h->size && cmp_str_id(h->a[l], h->a[i]) > 0){
largest = l;
}
else{
largest = i;
}
if(r <= h->size && cmp_str_id(h->a[r], h->a[largest]) > 0){
largest = r;
}
return largest;
}
void max_heapify(heap* h, int i){
int largest = get_largest(h, i);
while(largest != i){
swap(&(h->a[i]), &(h->a[largest]));
i = largest;
largest = get_largest(h, i);
}
}
void build_max_heap(heap* h){
for(int i = (h->size / 2); i >= 1; i--){
max_heapify(h, i);
}
}
int extract_max(heap* h){
if(h->size < 1) return -1;
int max_num = h->a[1];
h->a[1] = h->a[h->size];
h->size -= 1;
max_heapify(h, 1);
return max_num;
}
void insert(heap* h, int key){
h->size += 1;
int i = h->size;
h->a[i] = key;
while(i > 1 && cmp_str_id(h->a[parent(i)], h->a[i]) < 0){
swap(&(h->a[i]), &(h->a[parent(i)]));
i = parent(i);
}
}
int main(){
scanf("%d %d", &n, &m);
getchar();
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
for(int k = 0; k < n; k++){
int id = get_id(j, k, i);
scanf("%c", &tower[id]);
if(tower[id] == '0') tower[id] = '\0';
}
getchar();
}
}
heap* h = (heap*) malloc(sizeof(heap)); h->size = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
int id = get_id(i, j, 0);
insert(h, id);
}
}
while(h->size != 0){
int max_id = extract_max(h);
int len_str = strlen(&(tower[max_id]));
printf("%s\n", &(tower[max_id]));
if(tower[max_id + len_str + 1] != '\0'){
insert(h, max_id + len_str + 1);
}
}
free(h);
}