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 }