UVA 784 Maze Exploration （DFS || 种子填充）

Problem Description

A maze of rectangular rooms is represented on a two dimensional grid as illustrated 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 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 b) Painted maze

Figure 1. Mazes of rectangular rooms

All rooms of the maze are equal sized with all walls 3 points wide and 1 point thick as illustrated

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, 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 is a text file structured as follows:

1. The first line contains a positive integer 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

by a separation line full of underscores (‘ ’). There are at most 30 lines and at most 80 characters in a

line for each maze. The program reads the mazes from the standard input.

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 program writes the painted mazes on the standard 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

题解：

发现UVa的题题目描述一向很吓人，题目描述很长，要耐心看。

此题其实很简单，属于很裸的DFS，是求联通块的，这题题还有个很好听的别名，种子填充，代码简单易懂，就不多说了~

#include <iostream>
#include <cstring>
using namespace std;

const int MAXW = 100;
const int MAXH = 200;
int cases, r, c, ty, tx;
int direct[4][2] = { {0,-1}, {0,1}, {-1,0}, {1,0} };
char pic[MAXH][MAXW];

void dfs(int y, int x) {
for (int i = 0; i < 4; ++i) {
int nexty = y + direct[i][0];
int nextx = x + direct[i][1];
if (nexty < 0 || nextx < 0 || nexty >= r || nextx >= c || pic[nexty][nextx] != ' ') {
continue;
}
pic[nexty][nextx] = '#';
dfs(nexty, nextx);
}
}

int main() {
while (scanf("%d%*c", &cases) != EOF) {
while (cases--) {
memset(pic, 0, sizeof(pic));
r = 0, c = 0;
bool getloc = false;
while (true) {
cin.getline(pic[r++], MAXW);
int len = strlen(pic[r - 1]);
if (len > c) c = len;
if (pic[r - 1][0] == '_') break;
if (!getloc) {
for (int i = 0; i < len; ++i) {
if (pic[r - 1][i] == '*') {
ty = r - 1; tx = i;
getloc = true;
break;
}
}
}

}
pic[ty][tx] = '#';
dfs(ty, tx);
for (int i = 0; i < r; ++i)
for (int j = 0; ; ++j) {
if (pic[i][j] == '\0') {
putchar('\n');
break;
} else {
putchar(pic[i][j]);
}
}
}
}
return 0;
}