Red and Black

最新推荐文章于 2021-02-28 18:02:50 发布

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本文介绍了一个使用深度优先搜索算法解决特定问题的方法，即在一个由黑色和红色瓷砖组成的矩形房间里，从一个黑色瓷砖出发，只能在黑色瓷砖上行走，计算可以到达的黑色瓷砖数量。

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

题解：深搜。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

char map[50][50];
int d[4][2] = {{0,-1},{0,1},{-1,0},{1,0}};
int n,m;
int sx,sy;
int ans;

void dfs(int x,int y)
{
	if(x < 0 || y < 0 || x >= n || y >= m || map[x][y] == '#')
	{
		return;
	}
	ans++;
	map[x][y] = '#';
	for(int i = 0;i < 4;i++)
	{
		dfs(x + d[i][0],y + d[i][1]);
	}
}

int main()
{
	while(scanf("%d%d",&m,&n) != EOF && (m + n) != 0)
	{
		getchar();
		for(int i = 0;i < n;i++)
		{
			for(int j = 0;j < m;j++)
			{
				scanf("%c",&map[i][j]);
				if(map[i][j] == '@')
				{
					sx = i;
					sy = j;
				}
			}
			getchar();
		}
		
		ans = 0;
		dfs(sx,sy);
		
		printf("%d\n",ans);
	}

	
	return 0;
}