关于BFS搜索的思想， 最简单的水题，不过深有体会啊。

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.InputThe 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 arenot 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)OutputFor 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

相当于遍历一个二维数组，然后在能走到的范围里面计数能到达的位置。 很简单，就不停的递归就可以了。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <cstdlib>#include <queue>#include <stack>using namespace std;int cont; int M,N,i,j,k,t;char a[25][25];bool b[25][25];void bfs(int i ,int j){    if(i>0&&a[i-1][j]=='.'&&b[i-1][j]==0)    {  if(b[i-1][j]==0)        cont++;        b[i-1][j]=1;        bfs(i-1,j);    }    if(j>0&&a[i][j-1]=='.'&&b[i][j-1]==0)    {        if(b[i][j-1]==0)        cont++;        b[i][j-1]=1;        bfs(i,j-1);    }    if(i<M-1&&a[i+1][j]=='.'&&b[i+1][j]==0)    {        if(b[i+1][j]==0)        cont++;        b[i+1][j]=1;        bfs(i+1,j);    }    if(j<N-1&&a[i][j+1]=='.'&b[i][j+1]==0)    {        if(b[i][j+1]==0)        cont++;        b[i][j+1]=1;        bfs(i,j+1);    }}int main(){   while(scanf("%d%d",&N,&M)!=EOF)   {       if(M==0&&N==0)        break;       cont=1;       getchar();       memset(b,0,sizeof(b));       for(i=0;i<M;i++)        {      for(j=0;j<N;j++)                {                    scanf("%c",&a[i][j]);                    if(a[i][j]=='@')                    {                        k=i;                        t=j;                        b[i][j]=1;              }         }              getchar();        }        bfs(k,t);        printf("%d\n",cont);   }return 0;}