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#include<iostream> | ||
#include<queue> | ||
using namespace std; | ||
int tom[101][101][101]; //토마토의 상태를 저장하기위한 배열 | ||
int col, row, height; | ||
int x[6] = { 0,0,-1,1,0,0 }; | ||
int y[6] = { -1,1,0,0,0,0 }; | ||
int z[6] = { 0,0,0,0,1,-1 }; | ||
struct tomato { //토마토의 좌표를 저장하기위한 구조체 | ||
int x1; //열 | ||
int y1; //행 | ||
int z1; //높이 | ||
tomato(int _x1, int _y1, int _z1) { | ||
x1 = _x1; y1 = _y1; z1 = _z1; | ||
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} | ||
}; | ||
queue<tomato>q; | ||
void bfs() { | ||
while (!q.empty()) { | ||
tomato t = q.front(); | ||
q.pop(); | ||
for (int i = 0; i < 6; i++) { //탐색범위(상,하,좌,우,위,아래) | ||
int _x = t.x1 + x[i]; | ||
int _y = t.y1 + y[i]; | ||
int _z = t.z1 + z[i]; | ||
if (_x < col && _x >= 0 && _y < row && _y >= 0 && _z < height && _z >= 0 && tom[_z][_y][_x] == 0) { | ||
tom[_z][_y][_x] = tom[t.z1][t.y1][t.x1] + 1; | ||
q.push(tomato(_x, _y, _z)); | ||
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} | ||
} | ||
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} | ||
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} | ||
int main(void) { | ||
ios_base::sync_with_stdio(false); | ||
cin.tie(NULL); | ||
cin >> col >> row >> height; | ||
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for (int i = 0; i < height; i++) { | ||
for (int j = 0; j < row; j++) { | ||
for (int k = 0; k < col; k++) { | ||
cin >> tom[i][j][k]; | ||
if (tom[i][j][k] == 1) //익은토마토가 들어있으면 | ||
q.push(tomato(k, j, i)); //탐색시작점을위해 그 좌표를 큐에 넣는다. | ||
} | ||
} | ||
} | ||
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bfs(); // | ||
int sum = 0; | ||
for (int i = 0; i < height; i++) { | ||
for (int j = 0; j < row; j++) { | ||
for (int k = 0; k < col; k++) { | ||
if (tom[i][j][k] == 0) //bfs후 안익은토마토가 있다면 모두익지못하는 상황이므로 -1을 출력 | ||
{ | ||
cout << -1; | ||
return 0; | ||
} | ||
if (sum < tom[i][j][k]) sum = tom[i][j][k]; //bfs 후 걸린시간을 찾기위해 tom배열의 최대값을 찾아 sum에 저장하는과정 | ||
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} | ||
} | ||
} | ||
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sum == 1 ? cout << 0 : cout << sum - 1; | ||
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} |