Roll The Cube C语言

Question

Problem Description
This is a simple game.The goal of the game is to roll two balls to two holes each.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up.
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.

Input
First there's an integer T(T<=100) indicating the case number.
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).

Output
The minimum times you press to achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.

Sample Input
4
6 3

---

B
B
H
H

---

4 4

---

BB
HH

---

4 4

---

BH
HB

---

5 6

---

.BB*
.H*H
...*

---

Sample Output
3
1
2
Sorry , sir , my poor program fails to get an answer.
Roll The Cube

Answer 1

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#include <stdio.h>
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int n, m;
        scanf("%d%d", &n, &m);
        char grid[n][m];
        for(int i = 0; i < n; ++i) {
            scanf(" %s", grid[i]);
        }
        int x[4] = {0, 0, -1, 1}, y[4] = {-1, 1, 0, 0};
        int res = 0x7fffffff;
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                int count = 0;
                if(grid[i][j] == 'B') {
                    for(int k = 0; k < 4; ++k) {
                        int nx = i + x[k], ny = j + y[k];
                        if(nx >= 0 && nx < n && ny >= 0 && ny < m && grid[nx][ny] != '*') {
                            ++count;
                        }
                    }
                    if(count == 2) {
                        res = min(res, count + 1);
                    }
                }
            }
        }
        printf("%d\n", res);
    }
    return 0;
}