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[][];

            m[][]=-; m[][]=;

            m[][]=; m[][]=;

            m[][]=; m[][]=-;

            m[][]=; m[][]=; /// 上下左右四种位移

        int h,w,flag[][];

        while(~scanf("%d%d",&h,&w)&&h&&w)

        {

            int a,b,sum=; char ch[][];

            for(int i=;i<=w;i++)

            {

                getchar();

                for(int j=;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,,sizeof(flag));

            flag[a][b]=;

            queue <int> x,y;

            x.push(a),y.push(b);

            while(!x.empty()&&!y.empty())

            {

                for(int i=;i<;i++)

                {

                    a=x.front()+m[i][];

                    b=y.front()+m[i][];

                    if(a>&&a<=w&&b>&&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]=;

                }

                x.pop(), y.pop();

            }

            printf("%d\n",sum);

        }

        return ;

}

巴特西

Red and Black 模板题 /// BFS oj22063

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