Hduoj2952【搜索水题】

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2156    Accepted Submission(s): 1421

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
   
   
   
   

 

Source

IDI Open 2009 

#include<stdio.h> 
int  sum, n, m, dx[4]={-1,1,0,0}, dy[4]={0,0,-1,1};
char map[110][110];
void dfs(int x, int y)
{
	map[x][y] = '.';
	for(int d = 0; d < 4; d++)
	{
		int nx=x+dx[d], ny=y+dy[d];
		if(nx>=0&&nx<n&&ny>=0&&ny<m&&map[nx][ny]!='.')
		{
			map[nx][ny] = '.';
			dfs(nx,ny);
		}
	}
}
int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d%d", &n, &m);
		for(int i = 0; i < n; i++)
		scanf("%s", map[i]);
		sum = 0;
		for(int i = 0; i < n; i++)
		{
			for(int j = 0; j < m; j++)
			{
				if(map[i][j] == '#')
				{
					dfs(i,j);
					sum++;
				}
			}
		}
		printf("%d\n", sum);
	}
}

题意：给出一个迷宫，由‘。’和‘#’组成，若‘#’的上下左右也是‘#’，那么就说这两个属于同一个块，现在问迷宫内有多少个块。

思路：就是基础的搜索题，用深搜去做，都不需要回溯，直接标记就是了。