算法5：递归与BFS、DFS

写在前面：本文记录一些贪心算法的练习题，欢迎讨论！本系列系本人原创，如需转载或引用，请注明原作者及文章出处。本文持续更新。

一、Red and Black (POJ1979)

描述 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.
输入 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)
The end of the input is indicated by a line consisting of two zeros.
输出 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). 样例输入

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

样例输出

代码

#include<iostream>
#include<algorithm>
using namespace std;
int w, h, beginx, beginy, arr[25][25], num;
char a[25][25];

void dfs(int x, int y)
{
	if (x > 0 && arr[x - 1][y] == 1)
	{
		num++;
		arr[x - 1][y] = 0;
		dfs(x - 1, y);
	}

	if (x < h - 1 && arr[x + 1][y] == 1)
	{
		num++;
		arr[x + 1][y] = 0;
		dfs(x + 1, y);
	}

	if (y > 0 && arr[x][y - 1] == 1)
	{
		num++;
		arr[x][y - 1] = 0;
		dfs(x, y - 1);
	}

	if (y < w - 1 && arr[x][y + 1] == 1)
	{
		num++;
		arr[x][y + 1] = 0;
		dfs(x, y + 1);
	}
}

int main()
{
	while (1)
	{
		cin >> w >> h;
		if (w == 0 && h == 0)
			break;
		for (int i = 0; i < h; i++)
			for (int j = 0; j < w; j++)
			{
				cin >> a[i][j];
				if (a[i][j] == '@')
				{
					beginx = i;
					beginy = j;
				}
				if (a[i][j] == '#' || a[i][j] == '@')
					arr[i][j] = 0;
				else
					arr[i][j] = 1;
			}

		num = 1;
		dfs(beginx, beginy);
		cout << num << endl;
	}
	return 0;
}

二、棋盘问题 (POJ1321)

描述在一个给定形状的棋盘（形状可能是不规则的）上面摆放棋子，棋子没有区别。要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列，请编程求解对于给定形状和大小的棋盘，摆放k个棋子的所有可行的摆放方案C。输入输入含有多组测试数据。
每组数据的第一行是两个正整数，n k，用一个空格隔开，表示了将在一个n*n的矩阵内描述棋盘，以及摆放棋子的数目。 n <= 8 , k <= n
当为-1 -1时表示输入结束。
随后的n行描述了棋盘的形状：每行有n个字符，其中 # 表示棋盘区域， . 表示空白区域（数据保证不出现多余的空白行或者空白列）。
输出对于每一组数据，给出一行输出，输出摆放的方案数目C （数据保证C<2^31）。 样例输入

2 1
#.
.#
4 4
...#
..#.
.#..
#...
-1 -1

样例输出

2
1

代码

#include<iostream>
#include<algorithm>
using namespace std;
int n, k, chess[10], num;
char a[10][10];

void dfs(int row, int k)
{
	if (k == 0)
	{
		num++;
		return;
	}

	for (int i = row; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (a[i][j] == '#' && chess[j] == 1)
			{
				chess[j] = 0;
				dfs(i + 1, k - 1);
				chess[j] = 1;
			}
		}
	}
}

int main()
{
	while (1)
	{
		cin >> n >> k;
		if (n == -1 && k == -1)
			break;

		for (int i = 0; i < n; i++)
			chess[i] = 0;

		num = 0;

		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
			{
				cin >> a[i][j];
				if (a[i][j] == '#')
					chess[j] = 1;
			}

		dfs(0, k);
		cout << num << endl;
	}
	return 0;
}

算法5：递归与BFS、DFS

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