hdu 1312 Red and Black

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
  

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

搜索入门；

#include<stdio.h>
#include<iostream>
using namespace std;
char dp[22][22];
int n,m;
int count;
int dx[4] = {0,1,0,-1};//代表方向  向下，向上，向左，向右 
int dy[4] = {1,0,-1,0};
bool judge(int x,int y)
{
     if(x<1||x>n||y<1||y>m)//到边界的话 
        return true;
     if(dp[x][y]=='#')//遇到不能走的话 
        return true;
      return false;
} 
void dfs(int a,int b)
{
    count++;//增加计数 
    dp[a][b] = '#';//走过则标记
    int i;
    for(i=0;i<4;i++)
     {
          int nx=a+dx[i];  //向四个方向搜索 
          int ny=b+dy[i];
          if(judge(nx,ny))
            continue;
         else
            dfs(nx,ny);  //该处可以走，继续搜索    
     } 
    
}
int main()
{
     while(scanf("%d%d",&m,&n)&&!(n==0&&m==0))
      {
          int xf;
          int yf;
        int i,j;
           for(i=1;i<=n;i++)
              {
                    for(j=1;j<=m;j++)
                      {
                         // scanf("%c",&dp[i][j]);  //不知道为什这样是错的 
                           cin>>dp[i][j];
                           if(dp[i][j]=='@') //找到起始位置
                           {
                              xf=i,yf=j; 
                              
                             }
                     }
                    getchar();
            }
              count=0;
              dfs(xf,yf);
              printf("%d\n",count);
      }
    return 0;
}