POJ - 1979 - Red and Black

Red and Black

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

输出。此人最多能走多少个黑格子.

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
char map[50][50];
bool vis[50][50];
int ans,h,l;
void dfs(int x,int y)
{
  for(int i=0;i<4;i++)
  {
    int fx=x+dx[i];
    int fy=y+dy[i];
    if(fx<0||fx>=h||fy<0||fy>=l)
    continue;
    if(vis[fx][fy]==0&&map[fx][fy]=='.')
    {
      //cout<<fx<<" "<<fy<<endl;
      ans++;
      vis[fx][fy]=1;
      dfs(fx,fy);
    }
  }
}
int main()
{
  while(cin>>l>>h)//先输入列 再输入行 
  {
    int x,y;
    if(h==0&&l==0)
    break;
    else 
    {
      for(int i=0;i<h;i++)
      {
        cin>>map[i];
      }
      for(int i=0;i<h;i++)
      {
        for(int j=0;j<l;j++)
        {
          if(map[i][j]=='@')
          {
            x=i;
            y=j;
          }
        }
      }
      memset(vis,0,sizeof(vis));
      ans=1;//@也算
      vis[x][y]=1;
      dfs(x,y);
      cout<<ans<<endl; 
    }
  }return 0;
 }