hdu 题目2952 Counting Sheep （DFS）

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1706    Accepted Submission(s): 1104

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

 

 简单dfs

 

#include<stdio.h>

int h,w;
char a[100][100];

void dfs(int x,int y){
    if(x-1>=0 && a[x-1][y]=='#')
    {    a[x-1][y]='.';    dfs(x-1,y);}
    if(y-1>=0 && a[x][y-1]=='#')
    {    a[x][y-1]='.';    dfs(x,y-1);} 
    if(x+1<h && a[x+1][y]=='#')
    {    a[x+1][y]='.';    dfs(x+1,y);} 
    if(y+1<w && a[x][y+1]=='#')
    {    a[x][y+1]='.';    dfs(x,y+1);} 
}

int main(){
    int t,i,j,num;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&h,&w);
        for(i=0;i<h;i++)    scanf("%s",a[i]);
        num=0;
        for(i=0;i<h;i++)
            for(j=0;j<w;j++){
                if(a[i][j]=='#'){
                    a[i][j]='.'; dfs(i,j);
                    num++;
                }
            }
        printf("%d\n",num);
    }
    return 0;
}