Red and Black POJ

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18311		Accepted: 9717

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

简单的dfs搜索：

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int vis[30][30];
char room[30][30];
int W,H;
void dfs(int x,int y)
{
    if(room[x][y]=='#'||vis[x][y]==1||x<0||y<0||x>=H||y>=W)
    return ;
    vis[x][y] = 1;
    dfs(x,y-1);
    dfs(x,y+1);
    dfs(x-1,y);
    dfs(x+1,y);
}
int main()
{
    while(scanf("%d%d",&W,&H)==2)
    {
        memset(vis,0,sizeof(vis));
        if(W==0&&H==0)
        break;
        for(int i = 0;i<H;++i)
        scanf("%s",room[i]);

        int a,b;
        for(int i = 0;i<H;++i)
         for(int j = 0;j<W;++j)
              if(room[i][j]=='@')
                {
                    a=i;
                    b=j;
                    break;
                }

        dfs(a,b);
        int count = 0;
        for(int i = 0;i<H;++i)
         for(int j = 0;j<W;++j)
           if(vis[i][j]==1)
           count++;
           printf("%d\n",count);

    }
}