POJ 1164 The Castle

The Castle
 
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Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6348		Accepted: 3592

Description

1   2   3   4   5   6   7

#############################

1 #   |   #   |   #   |   |   #

#####---#####---#---#####---#

2 #   #   |   #   #   #   #   #

#---#####---#####---#####---#

3 #   |   |   #   #   #   #   #

#---#########---#####---#---#

4 #   #   |   |   |   |   #   #

#############################

(Figure 1)

#  = Wall

|  = No wall

-  = No wall

Figure 1 shows the map of a castle.Write a program that calculates

1. how many rooms the castle has

2. how big the largest room is

The castle is divided into m * n (m<=50, n<=50) square modules. Each such module can have between zero and four walls.

Input
Your program is to read from standard input. The first line contains the number of modules in the north-south direction and the number of modules in the east-west direction. In the following lines each module is described by a
number (0 <= p <= 15). This number is the sum of: 1 (= wall to the west), 2 (= wall to the north), 4 (= wall to the east), 8 (= wall to the south). Inner walls are defined twice; a wall to the south in module 1,1 is also indicated as a wall to the north in
module 2,1. The castle always has at least two rooms.
Output
Your program is to write to standard output: First the number of rooms, then the area of the largest room (counted in modules).
Sample Input

4
7
11 6 11 6 3 10 6
7 9 6 13 5 15 5
1 10 12 7 13 7 5
13 11 10 8 10 12 13

Sample Output

5
9

 
 
//POJ 1164

#include <stdio.h>

int r, c, p[50][50], rooms, max, modules;

//r,c：南北向、东西向的方块数

//p[50][50]：输入的每个方块的数字

//rooms：城堡的房间数

//max：最大房间的面积

//modules：当前房间的面积

bool visit[50][50];  //标识房间是否被访问过

void markRoom(int x,int y);//搜寻与(x,y)属于同一房间的方块

int main()

{

 while(scanf("%d %d",&r,&c)==2)  //南北向，东西向的方块数

    {

  rooms=max=0;        //初始化房间数和最大房间的面积

  int i,j;

  for(i=0;i<r;i++)  for(j=0;j<c;j++)  

        {

   scanf("%d",&p[i][j]);

   visit[i][j]=0;   //输入每个房间的墙数，并且将本房间初始化为0（bool）

  }

  for(i=0;i<r;i++) for(j=0;j<c;j++)

        {

   modules=0;

   markRoom(i,j);

   if(modules!=0)

    rooms++;

   if(modules>max)

    max=modules;

  }

  printf("%d\n%d\n",rooms,max);

 }

 return 0;

}

void markRoom(int x,int y)

{

 if(visit[x][y])    return;

 else

    {

  visit[x][y]=1;

  modules++;

 }

 if ( p[x][y]<8 )   markRoom(x+1, y);    //按照下面的注释一步一步找

 else    p[x][y] = p[x][y]%8; 

    if ( p[x][y]<4 )    markRoom(x, y+1);

 else   p[x][y] = p[x][y]%4; 

    if ( p[x][y]<2 )    markRoom(x-1, y);

 else   p[x][y] = p[x][y]%2;

 if ( p[x][y]==0 )   markRoom(x, y-1);

}

/*

p<8

 (i,j)没有南墙,(i+1,j)与(i,j)属于同一房间

所有与(i+1,j) 属于同一房间的方块也与(i,j)属于同一房间

p%8<4

 (i,j)没有东墙， (i,j+1)与(i,j)属于同一房间

所有与(i,j+1)属于同一房间的方块也与(i,j)属于同一房间

p%4<2

(i,j) 没有北墙 ，(i-1,j)与(i,j)属于同一房间

所有与(i-1,j)属于同一房间的方块也与(i,j)属于同一房间

p%2=0

 (i,j)没有西墙，(i,j-1)与(i,j)属于同一房间

所有与(i,j-1)属于同一房间的方块也与(i,j)属于同一房间

*/