알고리즘 연습/백준
[C++][토마토 BOJ7569] BFS
_h.j
2021. 8. 3. 09:34
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이번엔 3차원
2차원 토마토에서 축 하나 더 생긴거라 비슷하게 가면 된다.
소스코드
#include <bits/stdc++.h>
using namespace std;
int box[101][101][101]={0,};
int h,n,m;
int ans=0;
struct pos{
int x,y,z;
};
queue<pos> q;
int dz[6]={-1,1,0,0,0,0},dy[6]={0,0,0,0,1,-1},dx[6]={0,0,1,-1,0,0};
void Input(){
cin >>m >> n >> h;
for(int i = 0 ; i < h;i++){
for(int j = 0 ; j < n;j++){
for(int k = 0 ; k < m;k++){
cin >> box[i][j][k];
if(box[i][j][k]==0) numOfNotRipe++;
else if(box[i][j][k]==1){
q.push({i,j,k});
}
}
}
}
}
void Solution(){
while (!q.empty()) {
int x = q.front().x, y = q.front().y, z = q.front().z;
q.pop();
for (int i=0; i<6; i++) {
int nx = x + dx[i], ny = y + dy[i], nz = z + dz[i];
if (nx < 0 || nx >= h || ny < 0 || ny >= n || nz < 0 || nz >= m) continue;
if (box[nx][ny][nz]) continue; // Ripened(1 or above), empty(-1)
box[nx][ny][nz] = box[x][y][z] + 1; // Day count
q.push({nx, ny, nz});
}
}
for(int i = 0 ; i < h;i++){
for(int j = 0 ; j < n;j++){
for(int k = 0 ; k < m;k++){
if(box[i][j][k]==0) {
cout<<"-1";
return;
}
if (ans < box[i][j][k]) ans = box[i][j][k];
}
}
}
cout<< ans-1;
//numOfNotRipe ==0 ? cout<< ans : cout<< "-1";
}
void Slove(){
Input();
Solution();
}
int main(){
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
Slove();
}
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