hduoj 1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17813    Accepted Submission(s): 10846

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

4559613

题解：简单深搜；需要注意的是m,n是反的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char Map[30][30];
int vis[30][30];
int ans = 0;
int m, n,stx,sty;
int to[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
void dfs(int i,int j)
{
 /*if(i < 0||j < 0||i >= m||j >= n) return ;
 if(vis[i][j] == 1||Map[i][j] == '#') return ;
 vis[i][j] = 1;
 ans++;
 dfs(i-1,j);
 dfs(i+1,j);
 dfs(i,j+1);
 dfs(i,j-1);*/
 ans++; 
    Map[i][j] = '#'; 
    for(int k = 0; k<4; k++) 
    { 
        int x = i+to[k][0]; 
        int y = j+to[k][1]; 
        if(x<n && y<m && x>=0 && y>=0 && Map[x][y] == '.') 
            dfs(x,y); 
    } 

}
int main(){
 while(cin >> m >> n&&m)
 {
  memset(vis,0,sizeof(vis));
  ans = 0;
  for(int i = 0;i < n;i++)
  {
   for(int j = 0;j < m;j++)
   {
    cin >> Map[i][j];
     if(Map[i][j] == '@')
     {
      stx = i;
      sty = j;
     }
   }
  }
  dfs(stx,sty);
  cout << ans << endl;
 }
   return 0;

}