HD_1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14268    Accepted Submission(s): 8845

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
   
   
   
   

 

   简单搜索题，运用递归调用，对于每一个点向上下左右四个方向进行搜索，并对于搜索过的每一个点都做上标记以为了防止重复路径，然后整个路径搜索一遍之后的次数即为所要求的结果，搜索点的个数。代码如下：

<span style="font-size:14px;">include<stdio.h>

char map[25][25];
int  D[4][2]={{1,0},{0,1},{-1,0},{0,-1}};  //定义四个方向 
int  node,n,m;
	
void dfs(int i, int j){  
	node++; //记录每一次的搜索次数 
	map[i][j]='#';  //每搜索过一次做一个标记 
	for(int x=0; x<4; ++x){  //上下左右各搜索一次 
		int a=i+D[x][0];
		int b=j+D[x][1];
		if(a<m && a>=0 && b<n && b>=0 && map[a][b] == '.')  
			dfs(a,b);  	//递归搜索 
	}
	return;
}

int main(){
	while(scanf("%d%d", &n, &m)!=EOF){
		getchar();
		node=0;
		int di,dj;
		if(m == 0 && n == 0)	break;
		for(int i=0; i<m; ++i){
			for(int j=0; j<n; ++j){
				scanf("%c",&map[i][j]);
				if(map[i][j] == '@'){  //记录开始的位置 
					di=i;
					dj=j;
				}
			}
			getchar();
		}
		dfs(di,dj);
		printf("%d\n",node);
	}
return 0;
}</span>