Each tomato box is a three-dimensional grid. A ripe tomato ripens its unripe neighbors in the six axis-aligned directions, and all changes happen simultaneously one day at a time. The task is to find the number of days until no unripe tomatoes remain, or report -1 if some can never ripen.
Use multi-source breadth-first search: enqueue every initially ripe tomato before searching. Store each queued cell as one flattened integer index in a primitive int[] whose capacity is the total number of cells. A cell is enqueued only when it changes from unripe to ripe, so it enters the queue at most once.
The grid value doubles as the day label: initial ripe cells are 1, and a newly ripened cell gets its predecessor’s value plus one. Thus each BFS layer represents one simultaneous day. Count unripe cells while reading input; decrement the count as they ripen. If it starts at zero, the answer is 0. If it is still positive after the queue is exhausted, return -1; otherwise the largest grid value minus one is the number of elapsed days.
The input reader parses whitespace-separated integers without relying on line boundaries.
import java.io.BufferedInputStream;
import java.io.IOException;
public class Main {
private static int m, n, h;
private static final int[] DZ = {1, -1, 0, 0, 0, 0};
private static final int[] DY = {0, 0, 1, -1, 0, 0};
private static final int[] DX = {0, 0, 0, 0, 1, -1};
public static void main(String[] args) throws IOException {
FastScanner in = new FastScanner();
m = in.nextInt();
n = in.nextInt();
h = in.nextInt();
int volume = m * n * h;
int[][][] box = new int[h][n][m];
int[] queue = new int[volume];
int tail = 0;
int unripe = 0;
for (int z = 0; z < h; z++) {
for (int y = 0; y < n; y++) {
for (int x = 0; x < m; x++) {
int value = in.nextInt();
box[z][y][x] = value;
if (value == 1) {
queue[tail++] = (z * n + y) * m + x;
} else if (value == 0) {
unripe++;
}
}
}
}
System.out.println(ripeningDays(box, queue, tail, unripe));
}
private static int ripeningDays(int[][][] box, int[] queue, int tail, int unripe) {
if (unripe == 0) {
return 0;
}
int maxDay = 1;
for (int head = 0; head < tail; head++) {
int index = queue[head];
int z = index / (n * m);
int remainder = index % (n * m);
int y = remainder / m;
int x = remainder % m;
int nextDay = box[z][y][x] + 1;
for (int direction = 0; direction < 6; direction++) {
int nz = z + DZ[direction];
int ny = y + DY[direction];
int nx = x + DX[direction];
if (nz < 0 || nz >= h || ny < 0 || ny >= n || nx < 0 || nx >= m
|| box[nz][ny][nx] != 0) {
continue;
}
box[nz][ny][nx] = nextDay;
maxDay = nextDay;
unripe--;
queue[tail++] = (nz * n + ny) * m + nx;
}
}
return unripe == 0 ? maxDay - 1 : -1;
}
private static final class FastScanner {
private final BufferedInputStream in = new BufferedInputStream(System.in);
private final byte[] buffer = new byte[1 << 16];
private int pointer, length;
int nextInt() throws IOException {
int c;
do {
c = read();
} while (c <= ' ' && c != -1);
int value = 0;
while (c > ' ') {
value = value * 10 + c - '0';
c = read();
}
return value;
}
private int read() throws IOException {
if (pointer == length) {
length = in.read(buffer);
pointer = 0;
if (length == -1) {
return -1;
}
}
return buffer[pointer++];
}
}
}Every cell is inserted at most once and checks at most six neighbors, so the running time is O(MNH). The grid and primitive queue each use O(MNH) storage.