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[][];
 bool vis[][];
 int dx[]={,,,-},dy[]={,-,,};
 int w,h;
 int sx,sy;
 struct node
 {
     int x,y;
 };
 int bfs()
 {
     queue<node> q;
     int ans=;
     memset(vis,,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=;i<;i++)
         {
             temp.x=s.x+dx[i];
             temp.y=s.y+dy[i];
             if(temp.x>=&&temp.x<h&&temp.y>=&&temp.y<w&&vis[temp.x][temp.y]==&&map[temp.x][temp.y]=='.')
             {

                 vis[temp.x][temp.y]=;
                 q.push(temp);

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

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