##### minesweeper uva 10189 solution
```Have you ever played Minesweeper? It's a cute little game which comes within a certain Operating
System which name we can't really remember. Well, the goal of the game is to find where are all the
mines within a MxN field. To help you, the game shows a number in a square which tells you how
many mines there are adjacent to that square. For instance, suppose the following 4x4 field with 2
mines (which are represented by an `*' character):
*...
....
.*..
....
If we would represent the same field placing the hint numbers described above, we would end up
with:
*100
2210
1*10
1110
As you may have already noticed, each square may have at most 8 adjacent squares```

## Technique to solve this problem

As you may have already noticed, each square may have at most 8 adjacent squares. Let us suppose a mine is at (i,j)  location. its all adjacent blocs would be

To check all 8 adjacent blocks we will use two helper arrays  int dirx[8] = { -1, -1, -1, 0, 0, 1, 1, 1 } and int diry[8] = { -1, 0, 1, -1, 1, -1, 0, 1 }.

• Find each mine and increments count of all valid adjacent squares blocks.
```#include<iostream>
#include<cstdio>
using namespace std;

#define N 101
#define M 101

char mines[N][M];
int omines[N][M];
int dirx[8] = { -1, -1, -1, 0, 0, 1, 1, 1 };
int diry[8] = { -1, 0, 1, -1, 1, -1, 0, 1 };

void solve(int u, int v, int n, int m) {
for (int i = 0; i < 8; i++) {
int x = u + dirx[i];
int y = v + diry[i];
if (x < 0 || y < 0 || x >= n || y >= m)
continue;
if (mines[x][y] == '.') {
omines[x][y]++;
}
}
}

int main() {
int n = 0;
int m = 0;
freopen("in.txt", "r", stdin);
int t = 0;
int Flag = 0;
while (scanf("%d%d", &n, &m)) {
if (n == 0 && m == 0) {
break;
}
t++;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> mines[i][j];
omines[i][j] = 0;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (mines[i][j] == '*') {
omines[i][j] = '*';
solve(i, j, n, m);
}
}
}
// output
if (Flag == 1)
cout << "\n";
Flag = 1;
cout << "Field #" << t << ":" << endl;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (omines[i][j] == 42)
cout << '*';
else
cout << omines[i][j];
}
cout << endl;
}
}
return 0;
}

```