# Mine Sweeper

This problem is a game to simulate Mine Sweeper. To solve this problem we first count what number each cell should hold given a configration of mines. Then we use flood fill methods to mark all cells that holds a number that is zero or is adjoined with one zero. Every time we perform the flood filling we count it as one click. At last we count cells that holds number which have not been marked.

import java.io.*;
import java.util.*;

public class Minesweeper {
static private int[] dx = {-1, -1, -1, 0, 1, 1, 1, 0};
static private int[] dy = {-1, 0, 1, 1, 1, 0, -1, -1};
private static void countMines(int N, char[][] grid) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (grid[i][j] != '.') continue;
int count = 0;
for (int k = 0; k < 8; k++) {
if (i + dx[k] >= 0 && j + dy[k] >= 0 && i + dx[k] < N && j + dy[k] < N) {
count += (grid[i + dx[k]][j + dy[k]] == '*') ? 1 : 0;
}
}
grid[i][j] = (char)('0' + count);
}
}
}
private static void fillBFS(int N, int x, int y, char[][] grid) {

while (!xq.isEmpty()) {
int i = xq.pollFirst();
int j = yq.pollFirst();

char c = grid[i][j];

grid[i][j] = '.';
if (c != '0') continue;

for (int k = 0; k < 8; k++) {
if (i + dx[k] >= 0 && j + dy[k] >= 0 && i + dx[k] < N && j + dy[k] < N) {
char nc = grid[i + dx[k]][j + dy[k]];
if (nc >= '0' && nc <= '8') {
}
}
}
}
}
private static int solve(int N, char[][] grid) {
countMines(N, grid);
int count = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (grid[i][j] == '0') {
fillBFS(N, i, j, grid);
count++;
}
}
}

for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (grid[i][j] >= '0' && grid[i][j] <= '8') count++;
}
}
return count;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int t = 0; t < T; t++) {
int N = in.nextInt();
char[][] grid = new char[N][N];
for (int i = 0; i < N; i++) {
String l = in.next();
for (int j = 0; j < N; j++) {
grid[i][j] = l.charAt(j);
}
}
System.out.printf("Case #%d: %d\n", t + 1, solve(N, grid));
}
}
}