Problem 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)

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

4559613

Source

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思路

平平无奇的一道简单bfs问题，只要每次广搜入队的时候都统计一次就好了，最后返回结果并输出

代码

#include
    
     using namespace std;int a,b,c,t;const int d[][2]={ {-1,0},{0,1},{1,0},{0,-1} };struct node{    int x;    int y;}st,ed;int n,m;char maps[21][21];bool judge(node x){    if(x.x<=m && x.x>=1 && x.y<=n && x.y>=1 && maps[x.x][x.y]=='.')        return true;    return false;}int bfs(node st){    queue
     
       q;    q.push(st);    maps[st.x][st.y] = '#';    node now,next;    int t = 0;    while(!q.empty())    {        now = q.front();        q.pop();        for(int i=0;i<4;i++)        {            next.x = now.x + d[i][0];            next.y = now.y + d[i][1];                       if(judge(next))            {                q.push(next);                t++;                maps[next.x][next.y] = '#';            }        }    }    return t+1;//起点也算}        int main(){    while(cin>>n>>m)    {        if(n==0 && m==0) break;        for(int i=1;i<=m;i++)            for(int j=1;j<=n;j++)            {                cin >> maps[i][j];                if(maps[i][j]=='@')                {                    st.x = i; st.y = j;                }            }        int ans = bfs(st);        cout << ans << endl;    }    return 0;}