HDU Counting Sheep

Counting Sheep

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 64   Accepted Submission(s) : 51

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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
#include<iostream>
#include<queue>
int n, m;
char a[101][101], d[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
using namespace std;
int main()
{
	int T; scanf_s("%d", &T);
	while (T--)
	{
		scanf_s("%d%d", &n, &m);
		for (int i = 0; i < n; i++)
			cin >> a[i];
		int sum = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
			{
				if (a[i][j] == '#')
				{
					sum++;
					queue<int> P, Q;
					P.push(i), Q.push(j);
					a[i][j] = '.';
					while (!P.empty())
					{
						int x = P.front(), y = Q.front();
						P.pop(); Q.pop();
						for (int i = 0; i < 4; i++)
						{
							int xx = x + d[i][0], yy = y + d[i][1];
							if (xx >= 0 && xx < n&&yy >= 0 && yy < m&&a[xx][yy] == '#')
							{
								a[xx][yy] = '.';
								P.push(xx); Q.push(yy);
							}
						}
					}
				}
			}
		printf("%d\n", sum);
	}
	return 0;
}