uva 784 - Maze Exploration

　　　　　　784 - Maze Exploration

A maze of rectangular rooms is represented on a two dimensional grid as illustrated in gure 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 gure 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 gure 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

gure 1b.

Input

The program input is a text le structured as follows:

1.

The rst line contains a positive integer which shows the number of mazes to be painted.

2.

The rest of the le contains the mazes.

The lines of the input le 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.

SampleInput

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

_____

SampleOutput

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

_____

 

#include <iostream>
#include <stack>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <stack>
#include <list>
#include <sstream>
#include <cmath>

using namespace std;

#define ms(arr, val) memset(arr, val, sizeof(arr))
#define mc(dest, src) memcpy(dest, src, sizeof(src))
#define N 85
#define INF 0x3fffffff
#define vint vector<int>
#define setint set<int>
#define mint map<int, int>
#define lint list<int>
#define sch stack<char>
#define qch queue<char>
#define sint stack<int>
#define qint queue<int>

char maze[N][N];
int dir[4][2] = {0, -1, -1, 0, 1, 0, 0, 1};

void dfs(int i, int j)
{
    maze[i][j] = '#';
    int ii, jj;
    for (int k = 0; k < 4; k++)
    {
        ii = i + dir[k][0];
        jj = j + dir[k][1];
        if (ii >= 0 && jj >= 0 && maze[ii][jj] && maze[ii][jj] == ' ')
        {
            dfs(ii, jj);
        }
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    getchar();
    while (t--)
    {
        int n = 0, i = 0, j = 0;
        while (gets(maze[n]), maze[n++][0] != '_');
        while (true)
        {
            if (maze[i][j])
            {
                if (maze[i][j] == '*')
                {
                    dfs(i, j);
                    break;
                }
            }
            else
            {
                i++;
                j = -1;
            }
            j++;
        }
        i = 0;
        do 
        {
            puts(maze[i]);
        } while (maze[i++][0] != '_');
    }
    return 0;
}