关于深搜具体的过程一点理解

这几天看深搜一直纠结

函数里面for循环递归，那分出去的分支是先进行一个还是同时进行，彼此之间有没有影响？

答案应该是先进行一个而且有影响。

以此题为例（poj3278）

Red and Black

Time Limit: 1000 ms / Memory Limit: 32768 kb

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).

#include <cstdio>
#include <iostream>
#include <string>

using namespace std;
int black=0;
int n, k;
char str[25][25];
int dir[4][2] = { 1,0,0,-1,0,1,-1,0 };
int step=0;

void bfs(int x,int y)
{
	int i, j;
	int xx, yy;
	for (i = 0;i < k;++i)
	{
		for (j = 0;j < n;++j)
		{
			printf("%c", str[i][j]);

		}
		printf("\n");
	}
	for (i = 0;i < 4;++i)
	{
		xx = x + dir[i][0];
		yy = y + dir[i][1];

		if (xx >= 0 && yy >= 0 && xx < k&&yy < n &&str[xx][yy] == '.')
		{
			str[xx][yy] = '#';
			black++;
			step += 3 - i;
			bfs(xx, yy);
			


		}
		else
			printf("false%d\n",step);
	}
}
int main(void)
{
	while (scanf("%d%d", &n, &k) != EOF)
	{
		if (n == 0 && k == 0)
			break;
		memset(str, 0, sizeof(str));
		int i, j;
		int dx, dy;
		for (i = 0;i < k;++i)
		{
			getchar();
			for (j = 0;j < n;++j)
			{
				scanf("%c", &str[i][j]);
			}
		}
		for (i = 0;i < k;++i)
		{
			for (j = 0;j < n;++j)
			{
				if (str[i][j] == '@')
				{
					dx = i;
					dy = j;
				}
			}
		}
		str[dx][dy] = '#';
		black++;
		bfs(dx, dy);
		
		cout << black << endl;
		black = 0;
	}
}

以下面为例，可见先是找到一个可行的点一直继续下去，如果走不动就输出false,（以下、左、右、上的顺序进行）直到最后输出了15个false，因为最后一个点是4个false，所以是先一条路走下去，然后反过头来继续前面的循环，因为数组已经变化了，所以都是false，输出了11个，这里可用定义的变量step证明。

3 3
@..
...
...
#..
...
...
#..
#..
...
#..
#..
#..
false6
false6
#..
#..
##.
false7
false7
#..
#..
###
false8
false8
false8
#..
#.#
###
false8
#..
###
###
false10
false10
false10
##.
###
###
false10
false10
###
###
###
false11
false11
false11
false11
false11
false11
false11
false11
false11
false11
false11
false11
false11
false11
false11

而如何遍历所有的可能走法呢？

bfs(xx,yy)后把str[xx][yy]='.'就可以了。

Ep:

2 2

@.
..
#.
..
#.
#.
false3
false3
#.
##
false4
false4
false4
##
##
false4
false4
false4
false4
false4
false4
##
..
##
.#
false8
##
##
false10
false10
false10
false10
false10
false10
false10
false10
false10
false10
7