Red and Black 模板题 /// BFS oj22063

题目大意：

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 Hare 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

#include <bits/stdc++.h>
using namespace std;
int main()
{
            int m[4][2];
            m[0][0]=-1; m[0][1]=0;
            m[1][0]=1; m[1][1]=0;
            m[2][0]=0; m[2][1]=-1;
            m[3][0]=0; m[3][1]=1; /// 上下左右四种位移

        int h,w,flag[25][25];
        while(~scanf("%d%d",&h,&w)&&h&&w)
        {
            int a,b,sum=1; char ch[25][25];
            for(int i=1;i<=w;i++)
            {
                getchar();
                for(int j=1;j<=h;j++)
                {
                    scanf("%c",&ch[i][j]);
                    if(ch[i][j]=='@')    
                        a=i,b=j;  /// 记录人的位置，即起点
                }
            }

            memset(flag,0,sizeof(flag));
            flag[a][b]=1;

            queue <int> x,y;
            x.push(a),y.push(b);
            while(!x.empty()&&!y.empty())
            {
                for(int i=0;i<4;i++)
                {
                    a=x.front()+m[i][0];
                    b=y.front()+m[i][1];
                    if(a>0&&a<=w&&b>0&&b<=h //如果点在范围内，且
                       &&ch[a][b]!='#'&&!flag[a][b])//位于没走过的黑瓷砖
                        x.push(a), y.push(b), sum++; //放入队列 否则忽略
                    flag[a][b]=1;
                }
                x.pop(), y.pop();
            }
            printf("%d
",sum);
        }

        return 0;
}