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.

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.

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

2

4 4

#.#.

.#.#

#.##

.#.#

3 5

###.#

..#..

#.###

6

3

IDI Open 2009

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求#的连通块的数量，直接dfs

#include<iostream>

#include<cmath>

using namespace std;

char map[105][105];

int m, n, t;

int dir[4][2] = { { -1, 0 },{ 0, -1 }, { 0, 1 },{ 1, 0 } };

void dfs(int si, int sj)

{

    if (si <= 0 || sj <= 0 || si > m || sj > n)

        return;

    for (int i = 0; i < 4; i++)

    {

        if (map[si + dir[i][0]][sj + dir[i][1]] != '.')

        {

            map[si + dir[i][0]][sj + dir[i][1]] = '.';

            dfs(si + dir[i][0], sj + dir[i][1]);

        }

    }

    return;

}

int main()

{

    int o;

    while(~scanf("%d",&o))

    {

        while(o--)

        {

            cin >> m >> n;

            {

                for (int i = 1; i <= m;i++)

                    for (int j = 1; j <= n; j++)

                        cin >> map[i][j];

                t = 0;

                for (int i = 1; i <= m; i++)

                    for (int j = 1; j <= n; j++)

                    {

                        if (map[i][j] == '#')

                        {

                            map[i][j] = '.';

                            t++;

                            dfs(i, j);

                        }

                    }

                    printf("%d\n", t);

            }

        }

    }

        return 0;

    }