算法题 深/广搜-07-Counting Sheep

A while ago I had trouble sleeping. I used to lie awake, staring at the ceiling, for hours and hours. Then one day my grandmother suggested I tried counting sheep after I’d gone to bed. As always when my grandmother suggests things, I decided to try it out. The only problem was, there were no sheep around to be counted when I went to bed.
Creative as I am, that wasn’t going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while . is grass (or whatever you like, just not sheep). To make the counting a little more interesting, I also decided I wanted to count flocks of sheep instead of single sheep. Two sheep are in the same flock if they share a common side (up, down, right or left). Also, if sheep A is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps A and C are in the same flock.

Now, I’ve got a new problem. Though counting these sheep actually helps me fall asleep, I find that it is extremely boring. To solve this, I’ve decided I need another computer program that does the counting for me. Then I’ll be able to just start both these programs before I go to bed, and I’ll sleep tight until the morning without any disturbances. I need you to write this program for me.
Input
The first line of input contains a single number T, the number of test cases to follow.

Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. Then follows H lines, each containing W characters (either # or .), describing that part of the grid.
Output
For each test case, output a line containing a single number, the amount of sheep flock son that grid according to the rules stated in the problem description.

Notes and Constraints
0 < T <= 100
0 < H,W <= 100
Sample Input
2
4 4
#.#.
.#.#
#.##
.#.#
3 5
###.#
…#…
#.###
Sample Output
6
3

思路：求#的区域个数，用深搜即可查找完

#include<stdio.h>
char map[101][101];
int m, n;
int num;
void dfs(int x, int y) {
	map[x][y] = '.';
	int next_x, next_y;
	int k;
	int dir[4][2]{ { -1,0 },{ 1,0 },{ 0,-1 },{ 0,1 } };
	for (k = 0; k < 4; k++) {
		next_x = x + dir[k][0];
		next_y = y + dir[k][1];
		if (0 <= next_x && next_x < m && 0 <= next_y && next_y < n&&map[next_x][next_y] == '#') {
			dfs(next_x, next_y);
		}
	}
	

}
int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d %d", &m, &n);
		int number = 0;
		int i, j;
		for (i = 0; i < m; i++) {
				scanf("%s", map[i]);
		}
		for (i = 0; i < m; i++) {
			for (j = 0; j < n; j++) {
				if (map[i][j] == '#') {
					dfs(i, j);
					number++;
				}
			}
		}
		printf("%d\n", number);
	}
	return 0;
}

来源：CSDN

作者：实在不知道什么

链接：https://blog.csdn.net/weixin_43813718/article/details/104123264