uva 784 Maze Exploration

A maze(迷宫) of
rectangular(矩形的) rooms is represented on a two
dimensional(空间的) grid as
illustrated(阐明) in figure 1a. Each point of the grid is represented by a character. The points of room walls are marked by the same character which can
be any printable(印得出的) character different than `*‘, `_‘ and space. In figure 1 this character is `X‘. All the other points
of the grid are marked by spaces.

               XXXXXXXXXXXXXXXXXXXXX             XXXXXXXXXXXXXXXXXXXXX
               X   X   X   X   X   X             X###X###X###X   X   X
               X           X   X   X             X###########X   X   X
               X   X   X   X   X   X             X###X###X###X   X   X
               XXXXXX XXX XXXXXXXXXX             XXXXXX#XXX#XXXXXXXXXX
               X   X   X   X   X   X             X   X###X###X###X###X
               X   X     *         X             X   X###############X
               X   X   X   X   X   X             X   X###X###X###X###X
               XXXXXXXXXXXXXXXXXXXXX             XXXXXXXXXXXXXXXXXXXXX

a) Initial
maze(迷宫)                    b) Painted maze

Figure 1. Mazes of
rectangular(矩形的) rooms

All rooms of the maze are equal sized with all walls 3 points wide and 1 point thick as
illustrated(阐明) in figure 2. In addition, a wall is shared on its full length by the separated rooms. The rooms can communicate through doors, which
are positioned in the middle of walls. There are no outdoor doors.

                     door
                       |
                     XX XX
                     X . X   measured from within the room
               door - ...--  walls are 3 points wide
                     X . X__
                     XXXXX  |
                       |___  walls are one point thick

Figure 2. A room with 3 doors

Your problem is to paint all rooms of a maze which can be visited starting from a given room, called the `start room‘ which is marked by a star (`*‘) positioned in the middle of the room. A room can be visited from another room if there is a door on
the wall which separates the rooms. By
convention(大会), a room is painted if its entire surface, including the doors, is marked by the character `#‘ as shown in figure 1b.

Input

The program
input(投入) is a text file
structured(有结构的) as follows:

1.

The first line contains a positive(积极的)
integer(整数) which shows the number of mazes to be painted.

2.

The rest of the file contains the mazes.

The lines of the input file can be of different length. The text which represents a maze is
terminated(终止) by a separation line full of
underscores(底线) (`_‘). There are at most 30 lines and at most 80 characters in a line for each
maze(迷宫)

The program reads the mazes from the
input(投入) file, paints them and writes the painted mazes on the standard
output(输出).

Output

The output text of a painted maze has the same format as that which has been read for that maze, including the separation lines. The example below
illustrates(阐明) a simple input which contains a single maze and the corresponding output.

Sample Input

2
XXXXXXXXX
X   X   X
X *     X
X   X   X
XXXXXXXXX
X   X
X   X
X   X
XXXXX
_____
XXXXX
X   X
X * X
X   X
XXXXX
_____

Sample Output

XXXXXXXXX
X###X###X
X#######X
X###X###X
XXXXXXXXX
X   X
X   X
X   X
XXXXX
_____
XXXXX
X###X
X###X
X###X
XXXXX
_____

题目大意：将可以走到的地方标记成‘#‘(下划线是终止一组数据）

解题思路：DFS遍历的同时直接修改map.

#include<stdio.h>
#include<string.h>
char map[35][85], cnt = 0;
void DFS(int a, int b) {
	map[a][b] = '#';
	for (int i = -1; i <= 1; i++) {
		for (int j = -1; j <= 1; j++) {
			if (i == 0 && j == 0) {
				continue;
			}
			if (a + i < 0 || a + i > cnt) {
				continue;
			}
			if (b + j < 0 || b + j > strlen(map[a + i])) {
				continue;
			}
			if (map[a + i][b + j] == '#' || map[a + i][b + j] == 'X') {
				continue;
			}
			DFS(a + i, b + j);
		}
	}
}
int main() {
	int n;
	scanf("%d\n", &n);
	while (n--) {
		memset(map, 0, sizeof(map));
		cnt = 0;
		while (gets(map[cnt]) != NULL) {
			if (strcmp(map[cnt], "_____") == 0) {
				break;
			}
			cnt++;
		}
		for (int i = 0; i <= cnt; i++) {
			for (int j = 0; j < strlen(map[i]); j++) {
				if (map[i][j] == '*') {
					DFS(i, j);
				}
			}
		}
		for (int i = 0; i <= cnt; i++) {
			puts(map[i]);
		}
	}
	return 0;
}