hdu2952——Counting 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

Sample Input
2
4 4
#.#.
.#.#
#.##
.#.#
3 5
###.#
..#..
#.###

Sample Output
6
3分析：

四個方向暴搜；

源碼：

 
#include<stdio.h>   
#include<algorithm>   
#include<iostream>   
#include<string.h>   
using namespace std;  
int m,n;  
char map[120][120];  
int i,j;  
void dfs(int i,int j)  
{  
    if(map[i][j]!='#'||i<0||j<0||i>=m||j>=n)  
    return ;  
    else  
    {  
        map[i][j]='.';  
        dfs(i,j+1);  
        dfs(i,j-1);  
        dfs(i+1,j);  
        dfs(i-1,j);  
    }  
}  
int main()  
{  
    int tase;  
    scanf("%d",&tase);  
    while(tase--)  
    {  
        memset(map,0,sizeof(map));  
        scanf("%d%d",&m,&n);  
        for(i=0; i<m; i++)  
            for(j=0; j<n; j++)  
                cin>>map[i][j];  
        int p=0;  
        for(i=0; i<m; i++)  
            for(j=0; j<n; j++)  
            {  
                if(map[i][j]=='#')  
                {  
                    dfs(i,j);  
                    p++;  
                }  
            }  
            printf("%d\n",p);  
    }  
    return 0;  
}  

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
int m,n;
char map[120][120];
int i,j;
void dfs(int i,int j)
{
    if(map[i][j]!='#'||i<0||j<0||i>=m||j>=n)
    return ;
    else
    {
        map[i][j]='.';
        dfs(i,j+1);
        dfs(i,j-1);
        dfs(i+1,j);
        dfs(i-1,j);
    }
}
int main()
{
    int tase;
    scanf("%d",&tase);
    while(tase--)
    {
        memset(map,0,sizeof(map));
        scanf("%d%d",&m,&n);
        for(i=0; i<m; i++)
            for(j=0; j<n; j++)
                cin>>map[i][j];
        int p=0;
        for(i=0; i<m; i++)
            for(j=0; j<n; j++)
            {
                if(map[i][j]=='#')
                {
                    dfs(i,j);
                    p++;
                }
            }
            printf("%d\n",p);
    }
    return 0;
}