搜索入门-----HDU1312-优快云博客

本文介绍了一个基于深度优先搜索(DFS)的迷宫寻路算法，该算法可以在一个由黑色和红色方块组成的矩形网格中找到从指定起点可达的所有黑色方块。通过递归方式遍历所有可达路径，避免重复访问和无效路径，最终输出可到达的黑色方块总数。

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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

大致意思是： @是这个人的坐标，可以走 ' . ' , 但不能走’ # ‘, 问一共多少走法

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[25][25];
int sum, a, b;
void dfs(int m, int n){
	if(s[m][n] == '#')//遇见 # 返回 
	return ;
	if(m >= b || m < 0 || n >= a || n < 0)//超出范围返回 
	return ;
	sum++;
	s[m][n] = '#';//走过的不能再走 
	dfs(m + 1, n);
	dfs(m, n + 1);
	dfs(m - 1, n);
	dfs(m, n - 1);
}
int main()
{
	int m, n;
	while(scanf("%d%d", &a, &b) == 2 && a && b){
		int ok = 0;
		for(int i = 0; i < b; i++){
			scanf("%s", s[i]);
			if(!ok){
				for(int j = 0; j < a; j++){
					if(s[i][j] == '@'){
						m = i;
						n = j;
						ok = 1;
					}
				}
			}
		}
		sum = 0;
		dfs(m, n);
		printf("%d\n", sum);
	}
	return 0;
}