HDU 1312 Red and Black（DFS）

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

   
   
   
   

    
    
    
    45
59
6
13
   
   
   
   

 

//Must so
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<ctype.h>
#include<queue>
#include<vector>
#include<set>
#include<cstdio>
#include<cmath>
#define mem(a,x) memset(a,x,sizeof(a))
#define inf 1<<30
#define NN 1000006
using namespace std;
const double PI = acos(-1.0);
typedef long long LL;

/**********************************************************************
题意：
给出一个图，‘@’表示起点，‘.’表示可以走，‘#’表示不能走
只能走四个方向
求最多能走多少格（包括起点）
**********************************************************************/
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
bool vis[22][22];
char mp[30][30];
int row,cow;
int ans;
void dfs(int x,int y)
{

    for (int i = 0;i < 4;i++)
    {
        int xx = x + dx[i];
        int yy = y + dy[i];
        if (xx < 0||xx >= row||yy < 0||yy >= cow)
            continue;
        if (vis[xx][yy] == 0&&mp[xx][yy] == '.')
        {
            ans ++;
            vis[xx][yy] = 1;
            dfs(xx,yy);
        }
    }
}
int main()
{
    while (~scanf("%d%d",&cow,&row))//先输入行后输入列 
    {
    	if (row == 0&&cow == 0) break;
        int x,y;
        for (int i = 0;i < row;i++)
        {
           scanf("%s",mp[i]);
        }
        for (int i = 0;i < row;i++)
        {
        	for (int j = 0;j < cow;j++)
        	{
        		if (mp[i][j] == '@')
        		{
        			x = i;
        			y = j;
				}
			}
		}
        mem(vis,0);
        ans = 1;
        vis[x][y] = 1;
        dfs(x,y);
        printf("%d\n",ans);
    }
    return 0;
}