Counting Sheep
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3061 Accepted Submission(s): 2068
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.

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.
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
Notes and Constraints
0 < T <= 100
0 < H,W <= 100
Sample Input
2 4 4 #.#. .#.# #.## .#.# 3 5 ###.# ..#.. #.###
Sample Output
6 3
Source
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define N 105
int m,n;
int dir[4][2]= {{0,1},{1,0},{0,-1},{-1,0}};
int visited[N][N];
char map[N][N];
int sum=0;
void DFS(int x,int y)
{
int nx,ny;
for(int i=0; i<4; i++)
{
nx=x+dir[i][0];
ny=y+dir[i][1];
if(map[nx][ny]=='#'&&nx>=0&&nx<m&&ny>=0&&ny<n&&visited[nx][ny]==0)
{
visited[nx][ny]=1;
DFS(nx,ny);
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cin>>m>>n;
if(m==0&&n==0)
break;
for(int i=0; i<m; i++)
{
for(int j=0; j<n; j++)
cin>>map[i][j];
}
memset(visited,0,sizeof(visited));
sum=0;
for(int i=0; i<m; i++)
{
for(int j=0; j<n; j++)
{
if(map[i][j]=='#'&&visited[i][j]==0)
{
visited[i][j]=1;
DFS(i,j);
sum++;
}
}
}
cout<<sum<<endl;
}
return 0;
}