poj 1979 Red and Black（dfs水题）

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Japan 2004 Domestic

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 26
int n,m;
char mp[N][N];
int x,y;
int vis[N][N];
int ans;
int dirx[]={0,0,-1,1};
int diry[]={-1,1,0,0};
void dfs(int sx,int sy){
    for(int i=0;i<4;i++){
        int xx=sx+dirx[i];
        int yy=sy+diry[i];
        if(vis[xx][yy]) continue;
        if(mp[xx][yy]=='#') continue;
        if(xx<0 || xx>=n || yy<0 || yy>=m) continue;
        vis[xx][yy]=1;
        ans++;
        dfs(xx,yy);
    }
}
int main()
{
    while(scanf("%d%d",&m,&n)==2 && n+m){
        for(int i=0;i<n;i++){
            scanf("%s",mp[i]);
            for(int j=0;j<m;j++){
                if(mp[i][j]=='@'){
                    x=i;y=j;
                }
            }
        }
        memset(vis,0,sizeof(vis));
        ans=1;
        vis[x][y]=1;
        dfs(x,y);
        printf("%d
",ans);

    }
    return 0;
}