hdu2952Counting Sheep

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

bool mp[110][110];
char s[110];
bool visit[110][110];
int q[11000][2],l,r;
int d1[] = {1,-1,0,0};
int d2[] = {0,0,1,-1};


void bfs(int a,int b)
{
    l = r = 0;
    visit[a][b] = 1;
    q[r][0] = a,q[r++][1] = b;
    while(l<r)
    {
        int x = q[l][0],y = q[l++][1];
        for(int j = 0;j<4;j++)
        {
            int xx = x+d1[j],yy = y+d2[j];
            if(mp[xx][yy]&&!visit[xx][yy])
            {
                visit[xx][yy] = 1;
                q[r][0] = xx,q[r++][1] = yy;
            }
        }
    }
}


int main()
{
    int z,n,m,i,j,k;
    cin>>z;
    while(z--)
    {
        cin>>n>>m;
        memset(mp,0,sizeof(mp));
        memset(visit,0,sizeof(visit));
        for(i = 1;i<=n;i++)
        {
            scanf("%s",s);
            for(j = 0;j<m;j++)
            {
                mp[i][j+1] = (s[j] == '#')?1:0;
            }
        }
        int ans = 0;
        for(i = 1;i<=n;i++)
        {
            for(j = 1;j<=m;j++)
            {
                if(!visit[i][j]&&mp[i][j]) ans++,bfs(i,j);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}