简单搜索-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)

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

Code

#include<iostream>
#include <string.h>
#include <string>
#include <stdio.h>
using namespace std;


const int MAXN=1e3+1;
int map[MAXN][MAXN],l1,l2;
int d1[4]={0,0,-1,1},d2[4]={-1,1,0,0};


bool addable(int x,int y){
    return x>=0&&x<l1&&y>=0&&y<l2&&!map[x][y];
}

int ans=0;
void dfs(int sx,int sy){
    ans++;
    map[sx][sy]=1;
    for(int i=0;i<4;i++){
        int nx=sx+d1[i],ny=sy+d2[i];
        if(addable(nx,ny)){
            dfs(nx,ny);
        }
    }
}



int main(){
    //freopen("/Users/guoyu/Desktop/algorithm/in.txt", "r", stdin);
    while(scanf("%d%d",&l2,&l1)&&(l1*l1+l2*l2)){
        char c;
        int sx,sy;
        for(int i=0;i<l1;i++){
            for(int j=0;j<l2;j++){
                map[i][j]=0;
                //getchar();
                //scanf("%c",&c);
                cin>>c;
                if(c=='#') map[i][j]=1;
                if(c=='@'){
                    sx=i;
                    sy=j;
                }
            }

        }
        ans=0;
        dfs(sx,sy);
        printf("%d\n",ans);
    }
    return 0;
}