HDU2952:Counting Sheep(DFS)

Counting Sheep

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

Total Submission(s) : 35   Accepted Submission(s) : 24

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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<cstdio>
#include<cstring>
#include<deque>
using namespace std;

struct node
{
    int x,y;
};
int dr[][]={{,},{,},{-,},{,-}};
deque<node> s;
int i,j,n,m,num,t;
char ch[][];

void bfs(int x,int y)
{
    node t;
    t.x=x;
    t.y=y;
    ch[x][y]='*';
    s.push_back(t);
    while(!s.empty())
    {
        node p=s.front();
        for(int i=;i<;i++)
        {
            int xx=p.x+dr[i][];
            int yy=p.y+dr[i][];
            if (xx>= && xx<n && yy>= && yy<m && ch[xx][yy]=='#')
            {
                t.x=xx;
                t.y=yy;
                ch[xx][yy]='.';
                s.push_back(t);
            }
        }
        s.pop_front();
    }
    return;
}
int main()
{
    scanf("%d",&t);
    for(;t>;t--)
    {
       scanf("%d%d",&n,&m);
        num=;
        for(i=;i<n;i++)
            scanf("%s",&ch[i]);
        for(i=;i<n;i++)
            for(j=;j<m;j++)
             if (ch[i][j]=='#')
              {
                  bfs(i,j);
                  num++;
              }
      printf("%d\n",num);
    }
    return ;
}