Red and Black（红与黑）BFS

题目大意：

从 ‘  @  ’ 出发，只能走‘   .  ’ ,不能走‘ # ’，只能上下左右移动，最终可以走多少步，注意的是，@也算一步。

F - Red and Black

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

Submit Status

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 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)
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<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;

int x,y,ans,m,n;
char map[22][22];//地图 
bool vis[22][22];//表示访问过没有 
int move[4][2] = {0,1,0,-1,1,0,-1,0};//移动 其中a[i][0]表示X移动, a[i][1]表示y移动.
struct AC
{
	int x,y;
}node,temp;

bool judge (int x,int y)
{
	if (x>=0 && x<m && y>=0 && y<n )//判断坐标是否越界 
	return true;
	else
	return false;
}

int bfs(int x,int y)
{
	ans=0;记录答案 
	queue<AC>q;
	node.x = x;
	node.y = y;
	
	q.push(node);
	
	while (!q.empty())//模板写法 
	{
		AC top = q.front();
		q.pop();
		
		for (int i=0;i<4;i++)
		{
			temp.x = top.x + move[i][0];//分别上下左右走动 
			temp.y = top.y + move[i][1];
			if (judge(temp.x,temp.y)==true && vis[temp.x][temp.y]==false && map[temp.x][temp.y]=='.')
			{
				ans++;		//能走的话就加一下 
				q.push(temp);
				vis[temp.x][temp.y] = true;
			}
		}
	}
	return ans;
}


int main()
{

	while (cin>>n>>m)
	{
		ans=0;
		if (n==0 && m==0) break;
		memset(vis,0,sizeof(vis));
		
		getchar();
		for (int i=0;i<m;i++)
		scanf("%s",map[i]);
		
		for (int i=0;i<m;i++)
		{
			for (int j=0;j<n;j++)
			{
				if (map[i][j]=='@')//找到@的位置，作为bfs的参数 
				{
					x=i;
					y=j;
					break;
				}
			}
		}
		cout<<bfs(x,y)+1<<endl;//最终答案要加 1  
	}
	
	return 0;
}
//最简单的bfs模板题了，只要背了模板就能做出来