poj 1979 Red and Black - bfs

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

思路：
先找到，然后用bfs进行搜索，查找此点附近的四个点，如果是'.'并且之前没有走过这个点则把此点加入队列并且标记为'走过的点'，同时可以走的点计数加1。

源代码：

 1 #include <iostream>
 2 #include<queue>
 3 #include <stdio.h>
 4 #include<string.h>
 5 char s[25][25];
 6 int visit[25][25];
 7 using namespace std;
 8 int main()
 9 {
10 int h,w;
11    int x,y,a,b,i,j;
12     int sum=0;
13    while(scanf("%d %d",&w,&h))
14    {
15        if(w==0&&h==0) break;
16         memset(visit,0,sizeof(visit));
17        sum=0;
18        for(i=0;i<h;i++)
19           {
20           scanf("%s",s[i]);
21         
22            for(j=0;j<w;j++)
23             {
24                
25                 if(s[i][j]=='@')
26              {x=i;
27              y=j;
28              visit[i][j]=1;
29            
30              }
31              if(s[i][j]=='#')
32              visit[i][j]=1;
33             }
34           }
35    int ok;
36    queue<int>q;
37    q.push(x);
38    q.push(y);
39    while(!q.empty())
40    {ok=1;
41        a=q.front();
42        q.pop();
43        b=q.front();
44        q.pop();
45     if(a-1>=0&&s[a-1][b]=='.'&&visit[a-1][b]==0)
46     {
47        sum++;
48        visit[a-1][b]=1;
49        q.push(a-1);
50        q.push(b);
51       
52     }
53     if(a+1<h&&s[a+1][b]=='.'&&visit[a+1][b]==0)
54     {
55         sum++;
56         visit[a+1][b]=1;
57         q.push(a+1);
58         q.push(b);
59        
60     }
61     if(b-1>=0&&s[a][b-1]=='.'&&visit[a][b-1]==0)
62     {
63         sum++;
64         visit[a][b-1]=1;
65         q.push(a);
66         q.push(b-1);
67       
68     }
69     if(b+1<w&&s[a][b+1]=='.'&&visit[a][b+1]==0)
70     {
71         sum++;
72         visit[a][b+1]=1;
73         q.push(a);
74         q.push(b+1);
75        
76     }
77     for(i=0;i<h;i++)
78     {
79         for(j=0;j<w;j++)
80         {
81          
82         }
83     }
84   }
85   printf("%d\n",sum+1);
86  
87    }
88     return 0;
89 }