2013
11-10

# Red and Black

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.

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.

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).

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

45
59
6
13


import java.io.BufferedInputStream;
import java.util.Scanner;

/**
*
*poj1979
* 这道题是数据结构的第一道,刚开始是用c语言做的,但后来其他题目我都用了java去写,
* 所以现在只是简单把c语法转化为java法,统一一下编码而已
* @author NC
*/
public class Main {

private final static int MAXCOL = 22;
private final static int MAXROW = 22;
private static int[][] floor = new int[MAXROW][MAXCOL]; //最多是20行，20列，再加两行两列作为边界
private static int[][] visited = new int[MAXROW][MAXCOL]; //标记是否访问过，默认没有访问过

public static void main(String[] args) {

Scanner scan = new Scanner(new BufferedInputStream(System.in));
while (scan.hasNext()) {
int col = scan.nextInt();
int row = scan.nextInt();
int i = 0, j = 0, k = 0;
int count = 0;
int flag = 0;
char c;
//读取数据
if (col == 0 && row == 0) {
break;
}
//因为是循环输入，所以每一次都得初始化
for (i = 0; i < MAXROW; i++) {
for (j = 0; j < MAXCOL; j++) {
floor[i][j] = 0;
visited[i][j] = 0;
}
}
scan.nextLine();
//一个一个读取字符，将符号转换为数字
for (i = 1; i <= row; i++) {
char[] ss = scan.nextLine().trim().toCharArray();
for (j = 1; j <= col; j++) {
while (true) {
c = ss[j-1];
if (c == '.' || c == '#' || c == '@') {
break;
}
}
if (c == '.') {
floor[i][j] = 1;
} else if (c == '#') {
floor[i][j] = 2;
} else if (c == '@') {
floor[i][j] = 3;
}
}
}
//找出人所站的位置
flag = 0;
for (i = 1; i <= row; i++) {
for (j = 1; j <= col; j++) {
if (floor[i][j] == 3) {
flag = 1;
break; //要跳出两层
}
}
if (flag == 1) {
break;
}
}
//深度遍历
if (floor[i][j] == 3) {
DFS(i, j);
}
//count the number of tiles he can reach from the initial tile
count = 0;
for (i = 1; i <= row; i++) {
for (j = 1; j <= col; j++) {
if (visited[i][j] == 1) {
count++;
}
}
}
//print the number of tiles he can reach from the initial tile
System.out.println(count);
}
}

static void DFS(int i, int j) {
//标记访问过
visited[i][j] = 1;
//四个方向都要遍历，不能用else if
if (floor[i][j + 1] == 1 && visited[i][j + 1] == 0) {
DFS(i, j + 1);
}
if (floor[i][j - 1] == 1 && visited[i][j - 1] == 0) {
DFS(i, j - 1);
}
if (floor[i + 1][j] == 1 && visited[i + 1][j] == 0) {
DFS(i + 1, j);
}
if (floor[i - 1][j] == 1 && visited[i - 1][j] == 0) {
DFS(i - 1, j);
}
}
}

1. 一开始就规定不相邻节点颜色相同，可能得不到最优解。我想个类似的算法，也不确定是否总能得到最优解：先着一个点，随机挑一个相邻点，着第二色，继续随机选一个点，但必须至少有一个边和已着点相邻，着上不同色，当然尽量不增加新色，直到完成。我还找不到反例验证他的错误。。希望LZ也帮想想, 有想法欢迎来邮件。谢谢