Red and Black(poj 1979 bfs)

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 27891		Accepted: 15142

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

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
char map[25][25];
bool vis[25][25];
int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int w,h;
int sx,sy;
struct node
{
    int x,y;
};
int bfs()
{
    queue<node> q;
    int ans=0;
    memset(vis,0,sizeof(vis));
    node s,temp;
    s.x=sx,s.y=sy;
    q.push(s);
    while(!q.empty())
    {    
        s=q.front();
        q.pop();
        ans++;
        for(int i=0;i<4;i++)
        {
            temp.x=s.x+dx[i];
            temp.y=s.y+dy[i];
            if(temp.x>=0&&temp.x<h&&temp.y>=0&&temp.y<w&&vis[temp.x][temp.y]==0&&map[temp.x][temp.y]=='.')
            {
                
                vis[temp.x][temp.y]=1;    
                q.push(temp);

            }
        }
    }
    return ans;
}
int main()
{
    int i,j;
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&w,&h))
    {
        if(w==0&&h==0)
            break;
        for(i=0;i<h;i++)
            scanf("%s",map[i]);
        for(i=0;i<h;i++)
            for(j=0;j<w;j++)
                if(map[i][j]=='@')
                {
                    sx=i,sy=j;
                    break;
                }

        cout<<bfs()<<endl;
    }
    return 0;
}