Red and Black

陈小盈 2022-05-31

Red and Black

Time Limit: 1000MS

Memory Limit: 30000K

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

45
59
6
13

Source

Japan 2004 Domestic

java

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// package com.ccying.oj.poj;

import java.util.Scanner;

/**
 * <h2>Red and Black.</h2>
 *
 * @author ccying
 * @since 2022/5/31
 */
public class Mian {
    // 四个方位
    private final static int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    // 地图
    private static char[][] grids;
    // 是否被访问
    private static int[][] isVisit;

    private static void init(int row, int clo) {
        grids = new char[row][clo];
        isVisit = new int[row][clo];
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < clo; j++) {
                isVisit[i][j] = 0;
            }
        }
    }

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        while (cin.hasNext()) {
            // 输入
            int w = cin.nextInt();
            int h = cin.nextInt();
            if (w == 0 && h == 0) {
                break;
            }
            cin.nextLine(); // 去除换行符
            init(h, w);
            int x = 0;  // '@'行坐标
            int y = 0;  // '@'列坐标
            for (int i = 0; i < h; i++) {
                String inputStr = cin.nextLine();
                for (int j = 0; j < w; j++) {
                    grids[i][j] = inputStr.charAt(j);
                    if (grids[i][j] == '@') {
                        x = i;
                        y = j;
                    }
                }
            }

            // dfs
            int count = dfs(x, y);

            // 输出
            System.out.println(count);
        }
    }

    private static int dfs(int x, int y) {
        isVisit[x][y] = 1;
        int count = 1;
        for (int i = 0; i < 4; i++){
            int xx = x + dirs[i][0];
            int yy = y + dirs[i][1];
            if (xx < 0 || yy < 0 || xx >= grids.length || yy >= grids[0].length) {  // 非法出界
                continue;
            }
            if (grids[xx][yy] == '.' && isVisit[xx][yy] == 0) {
                count += dfs(xx, yy);
            }
        }
        return count;
    }
}
A + B Problem