杭电（hdu）ACM 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13453    Accepted Submission(s): 8331

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

 

经典的搜索题目，深度优先搜索，其描述如下：

void dfs()

{

   for(所有的邻接节点)

  {

      if(节点没有被遍历)

      {

          标记此节点；

          dfs(此节点);

      }

  }

}

此题代码如下：

#include 
#include 
#include 
#include 
using namespace std;

int w,c;
int p,q;
char arr[21][21];
int v[21][21];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int count=0;

void DFS(int x,int y)
{
    for(int i=0;i<=3;i++)
    {
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(xx>=0&&xx<c&&yy>=0&&yy<w&&arr[xx][yy]=='.'&&v[xx][yy]==0)
        {
            v[xx][yy]=1;
            count++;
            DFS(xx,yy);
        }
    }
}
int main()
{
    while(cin>>w>>c,w||c)
    {
        for(int i=0;i<c;i++)
           for(int j=0;j<w;j++)
           {
                   cin>>arr[i][j];
                   if(arr[i][j]=='@')
                   {
                       p=i;
                    q=j;
                }
                v[i][j]=0;
           }
        v[p][q]=1;
        count = 1;
        DFS(p,q);
        cout<<count<<endl;
    }
    return 0;
}