POJ 1979 Red and Black（DFS）

最新推荐文章于 2020-01-13 10:17:57 发布

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#算法竞赛入门经典 #UVa #DFS #算法

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本文介绍了一种基于深度优先搜索的迷宫寻路算法。该算法在一个由黑色和红色方块组成的矩形房间中找到所有可达的黑色方块数量。通过递归方式遍历每个相邻的黑色方块，并避免重复访问已探索过的区域。文章提供了完整的C++实现代码。

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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)
The end of the input is indicated by a line consisting of two zeros.

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

#include<iostream>  
using namespace std;

int m, n;
char c;
int tempx, tempy;
int cnt;
int flag[25][25];
char tiles[25][25];

int dfs(int x, int y)
{
	for (int i = -1; i < 2; i++)
	{
		for (int j = -1; j < 2; j++)
		{
			if ((i == 0 || j == 0) && (i != 0 || j != 0))
			{
				if (0 <= (x + i) && (x + i) < n && 0 <= (y + j) && (y + j) < m && flag[x + i][y + j] == 0 && tiles[x + i][y + j] == '.')
				{
					flag[x + i][y + j] = 1;
					cnt++;
					dfs(x + i, y + j);
				}
			}
		}
	}
	return 0;
}

int main() {
	

	while ((cin>>m>>n)&&(m!=0||n!=0))
	{
		cnt = 1;

		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{	
				cin >> c;
				flag[i][j] = 0;
				tiles[i][j] = c;
				if (c=='@')
				{
					tempx = i;
					tempy = j;
					flag[i][j] = 1;
				}
			}
		}

		dfs(tempx, tempy);
		cout << cnt << endl;
	}
	return 0;
}