Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 13

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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#

[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output 45 59 6 13

我理解的搜尋就是從一個起始點按某種規律對地圖上的每個可檢索點進行判斷（規律如：按前後左右走），直到滿足情況時跳出或全部檢索完畢；

下面是我這題的程式碼：

#include<iostream>
#include<string>
#include<string.h>
#include<sstream>
#include<queue>
using namespace std;
struct node
{
    int x,y;
}s;    //s為起點
int n,m,h;
char a[20][20];  //地圖
int dir[4][2]={{1,0},{0,1},{0,-1},{-1,0}};//走的方向
void bfs()
{
    queue<node>q;
    node t,x1;
    q.push(s);
    while (!q.empty())
    {
        t=q.front();
        q.pop();
        for (int i=0;i<4;i++)    //4個方向搜尋，滿足條件入隊
        {
            x1.y=t.y+dir[i][1];
            x1.x=t.x+dir[i][0];
            if(x1.x>=0&&x1.x<m&&x1.y>=0&&x1.y<n&&a[x1.x][x1.y]=='.')  
            {
                h++;
                a[x1.x][x1.y]='#';   //檢索的標記防止多次計算
                q.push(x1);
            }
        }
    }
    cout<<h<<endl;
}
int main()
{
    int i,j;
    while (cin>>n>>m,n+m)
    {
        h=1;
        for (i=0;i<m;i++)   //輸入地圖找到起點
            for (j=0;j<n;j++)
            {
                cin>>a[i][j];
                if (a[i][j]=='@')
                {
                    s.x=i;
                    s.y=j;
                }
            }
        bfs();
    }
    return 0;
}