hdu 2952 Counting Sheep

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3524    Accepted Submission(s): 2366

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>
using namespace std;
int n,m;
char a[105][105];
int d[4][2]={1,0,-1,0,0,1,0,-1};
int s;
bool safe (int x,int y)
{
    if(x>n||y>m||x<=0||y<=0) return 0;
    return 1;
}
void dfs(int x,int y)
{
    a[x][y]='.';
    int i;
    for(i=0;i<4;i++)
    {
        int xx=x+d[i][0];
        int yy=y+d[i][1];
        if(!safe(xx,yy))  continue;
        if(a[xx][yy]=='#')
        {
            a[xx][yy]='.';
            dfs(xx,yy);
        }
    }
}
int main()
{
    int T,i,j;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        s=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                cin>>a[i][j];
            }
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(a[i][j]=='#')
                {
                    s++;
                     dfs(i,j);
                }
            }
        }
        cout<<s<<endl;
    }
    return 0;
}