J - Red and Black_red and black abstract background with angled bloc-优快云博客

本文介绍了一个迷宫探索问题，通过深度优先搜索算法来计算人物在只能行走于黑色方块的迷宫中所能达到的最大范围。文章提供了一段C语言实现的代码示例，详细展示了如何读取迷宫布局并计算可达的黑色方块数量。

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

题意：一个人被困在一个迷宫中，只能走四个方向，求他能走的最大步数。

思路：读入迷宫，用深搜从起点出发，搜索这个点能走到的所有顶点即可。

#include<stdio.h>
#include<string.h>
char s[50][50];
int n,m;
int dfs(int x,int y)
{
	int i,j,k;
	if(x<0||x>n-1||y<0||y>m-1)//越界判断
		return 0;
	
	if(s[x][y]=='#')//判断是否是墙
	{
		return 0;
	}	
	else
	{
		s[x][y]='#';//不是墙时把走过的点进行修改
		return 1+dfs(x+1,y)+dfs(x-1,y)+dfs(x,y+1)+dfs(x,y-1);//搜索四个方向
	}
}
int main() 
{
	int i,j,k;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==0&&n==0)
			break;
		for(i=0;i<n;i++)
		{
			scanf("%s",s[i]);
		}
		int sum=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(s[i][j]=='@')//循环找起点
				{
					sum=dfs(i,j);	
				} 
			} 
		}
		printf("%d\n",sum);
	}
	return 0;
}