题目链接：传送门

Problem Description

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

题意：问#被.分为几个部分。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
char mp[101][101];
int a,b,s;
void dg(int i, int j)
{
    if(mp[i][j]!='#'||i<0||j<0||i>=a||j>=b)
        return;
    else
    {
        mp[i][j]='*';
        dg(i-1, j);
        dg(i, j-1);
        dg(i, j+1);
        dg(i+1, j);
    }
}
int main()
{
    int i,j;
    int w;
    cin>>w;
    while(w--)
    {
        cin>>a>>b;
        s=0;
        for(i=0;i<a;i++)
        {
            for(j=0;j<b;j++)
            {
                cin>>mp[i][j];
            }
        }
        for(i=0;i<a;i++)
        {
            for(j=0;j<b;j++)
            {
                if(mp[i][j]=='#')
                {
                    dg(i,j);
                    s++;
                }
            }
        }
        printf("%d\n",s);
    }
    return 0;
}