hdu1312 Red and Black

I - Red and Black

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d
& %I64u

Submit Status Practice HDU
1312

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

45
59
6
13

#include<stdio.h>
#include<queue>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m;
char a[25][25];
int b[25][25];
int dir[4][2]= {0,1,1,0,0,-1,-1,0};
struct node
{
    int x,y;
} w;
int sum=0;
void bfs(int x,int y)
{
    queue<node> q;
    int k,i,j;
    w.x=x;
    w.y=y;
    q.push(w);
    while(!q.empty())
    {
        node s=q.front();
        q.pop();
        for(int i=0; i<4; i++)
        {
            w.x=s.x+dir[i][0];
            w.y=s.y+dir[i][1];
            if(!b[w.x][w.y]&&a[w.x][w.y]!='#'&&w.x>=1&&w.x<=m&&w.y>=1&&w.y<=n)
            {
                sum++;
                b[w.x][w.y]=1;
                q.push(w);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m)&&(n||m))
    {
        sum=0;
        int i,j,x=0,y=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1; i<=m; ++i)
            scanf("%s",a[i]+1);
        for(i=1; i<=m; ++i)
            for(j=1; j<=n; ++j)
            {
                if(a[i][j]=='@')
                {
                    x=i;
                    y=j;
                    break;
                }
            }
        b[x][y]=1;
        bfs(x,y);
        printf("%d\n",sum+1);
    }
    return 0;
}

//#include<iostream>
//#include<string.h>
//#include<stdio.h>
//using namespace std;
//int direct[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1 };  /*定义方向， 左右，上下*/
//char map[21][21];               /*输入的字符串*/
//bool mark[21][21];              /*标记走过的路程*/
//bool flag;
//int W, H;
//int Dx, Dy;            //记录起始位置@，从这里开始进行搜索
//int ans;                //记录满足的个数。初始化为1，因为@也包含在内
///*****底下是核心算法，主要是从
//上下左右这四个方向进行搜索，注意
//满足搜索的条件是不能越界，不能是#，
//还有就是没有搜索过的--》主要是靠mark[i][j]
//来实现*******/
//void DFS( int x, int y )
//{
//    mark[x][y] = true;
//    for( int k = 0; k < 4; k ++ )
//    {
//        int p = x + direct[k][0];
//        int q = y + direct[k][1];
//        if( p >= 0 && q >= 0 && p < H && q < W && !mark[p][q] && map[p][q] != '#' )
//        {
//            ans ++;
//            DFS( p, q );
//        }
//    }
//    return;
//}
//
//int main()
//{
//    int i, j, k;
//    while( cin >> W >> H && ( W || H ) )   // W -> column, H -> row;
//    {
//        memset( mark, false, sizeof( mark ) );
//        for( i = 0; i < H; i ++ )
//            for( j = 0; j < W; j ++ )
//            {
//                cin >> map[i][j];
//                if( map[i][j] == '@' )
//                {
//                    Dx = i;
//                    Dy = j;
//                }
//            }
//        ans = 1;
//        DFS( Dx, Dy );
//        cout << ans << endl;
//    }
//}
//
//
//

//#include <iostream>
//
//using namespace std;
//char a[50][50];
//int m,n;
//int sum;
//void dfs(int i,int j){
//    a[i][j]='#';//把搜素到的点变成#
//    sum++;//sum用于计数每搜到一个点就记一下数
//    if(i-1>=0&&a[i-1][j]=='.')dfs(i-1,j);
//    if(i+1<m&&a[i+1][j]=='.')dfs(i+1,j);
//    if(j-1>=0&&a[i][j-1]=='.')dfs(i,j-1);
//    if(j+1>=0&&a[i][j+1]=='.')dfs(i,j+1);//四个方向搜素
//}
//int main()
//{
//    while(cin>>n>>m&&(m+n)){
//        for(int i=0; i<m; i++)
//        cin>>a[i];
//        sum=0;
//        for(int i=0; i<m; i++)
//        for(int j=0; j<n; j++){
//            if(a[i][j]=='@')
//                dfs(i,j);
//        }
//        cout<<sum<<endl;
//    }
//    return 0;
//}