hdoj1312Red and Black

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
    
    
    
    
代码：
    
    
    
    
#include<stdio.h> #include<string.h> char map[1010][1010]; int vis[1010][1010],n,m,ans,sx,sy; void dfs(int x,int y) {  if(x<0||x>=m||y<0||y>=n||vis[x][y]||map[x][y]=='#')  return ;  vis[x][y]=1;  ans++;  dfs(x+1,y);  dfs(x,y+1);  dfs(x-1,y);  dfs(x,y-1); } int main() {  while(scanf("%d%d",&n,&m)&&n||m)  {  ans=0;  memset(vis,0,sizeof(vis));  memset(map,0,sizeof(map));  for(int i=0;i<m;i++)  {  scanf("%s",&map[i]);    for(int j=0;j<n;j++)  {  if(map[i][j]=='@')  sx=i,sy=j;  }  }  dfs(sx,sy);  printf("%d\n",ans);  }  return 0; } 
    
    
    
    
思路：深搜，dfs，这是第一次接触深搜题，乍一看一点思路也没有啊，然后学长讲完之后也是晕头转向的，迷迷糊糊，晕。然后自己照着代码敲了几遍，记住代码，再根据代码慢慢看懂了，memset(vis,0,sizeof(vis)); memset(map,0,sizeof(map));这个是要记住的，就是把所有的行行竖竖清零，在dfs，bfs里经常用到。然后就是定义dfs了，要记住确定方向dfs(x+1,y);dfs(x,y+1);dfs(x-1,y);dfs(x,y-1);，分别代表上下左右，其次就是找原始的点，然后本题就ac啦，注意不会多多敲几遍哈！！