基础的深度优先搜索题目

通过对每一个位置的上下左右位置进行递归搜索。

题目描述：

Red and Black

Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ – a black tile

‘#’ – a red tile

‘@’ – a man on a black tile(appears exactly once in a data set)

The end of the input is indicated by a line consisting of two zeros.

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

#include<iostream>
#include<algorithm>
#include<fstream>
#include<string>

using namespace std;
void solve();
void dfs(int row, int col);
int result = 0;
string grid[10001];
int n=-1, m=-1;
int main(){

	while (true){
		cin >> n >> m;
		if (n == 0 && m == 0)break;

		for (int i = 0; i < m; i++)
			grid[i] = "";
		
		
		for (int i = 0; i < m; i++)
				cin >> grid[i];

		result = 0;
		solve();
		cout << result << endl;


	}
	return 0;
}

void solve(){
	int s_row=0, s_col=0;
	bool isfind = false;
	
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n;j++){
			if (grid[i][j] == '@'){
				s_row = i;
				s_col = j;
				isfind = true;
				break;
			}
			if (isfind)
				break;
	}
		
	}

	dfs(s_row, s_col);
	
	return;
}


void dfs(int row, int col){
	
		result++;
		int x, y;
		grid[row][col]='#';
	
//	int x, y;
	for (int dy=-1;dy<=1;dy++)
	for (int dx = -1; dx <= 1; dx++){
		for (int dy = -1; dy <= 1; dy++){
           //if x,y int the grid
			x=row+ dx;
			y=col+dy;
			if ((dx==0||dy==0)&&x>= 0 && x<m&&y>=0 && y < n&&grid[x][y]=='.')
				dfs(x, y);
		}
	}
	return;
}