SDNU-DFS——C - Red and Black

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)
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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6

13

dfs题

首先得根据输入找到起点

然后用dfs从起点开始找

做好界的判断

#include
#include
#include
#include
using namespace std;
const int maxx = 20 + 10;
char map[maxx][maxx];
int m , n , cnt;
int step[4][2]={1,0,0,1,-1,0,0,-1};

void dfs(int i,int j)  
{  
    cnt++;  
    map[i][j] = '#';  
    for(int k = 0; k<4; k++)  
    {  
        int x = i+step[k][0];  
        int y = j+step[k][1];  
        if(x<=n && y<=m && x>=1 && y>=1 && map[x][y] == '.')  
        dfs(x,y);  
    }  
    return;  
}  

int main()
{
	int i , j , mani , manj ;  
    while(scanf("%d%d", &m, &n), m + n)  
    {  
    
        for(i = 1; i<=n; i++)  
        {  
            for(j = 1; j<=m; j++)  
            {  
                cin >> map[i][j];  
                if(map[i][j] == '@')  
                {  
                    mani = i;  
                    manj = j;  
                }  
            }   
        }  
        cnt = 0;  
       
        dfs(mani,manj);  
        cout<