POJ 1164

The Castle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5557		Accepted: 3129

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

 1 //双重墙，只算一次 

 2 #include <iostream>

 3 #include <cstring>

 4 using namespace std;

 5 int map[55][55];

 6 bool vis[55][55];

 7 int area = 0,ans = 0;

 8 int col,row;

 9 void dfs(int i,int j)

10 {

11     if(i<1||j<1||i>row||j>col)

12         return ;

13     if(vis[i][j])

14         return ;

15     else

16     {

17         vis[i][j] = true;

18         ++ans;

19         //可能好几个方向均没墙，所以不是级联式if语句 

20         if(map[i][j]%2 == 0) // 1 2 4 8 西边肯定没墙 

21         {

22             if(j>=2)

23                 dfs(i,j-1);

24         }

25         map[i][j] /= 2;

26         if(map[i][j]%2 == 0) // 0 1 2 4 北边肯定没墙 

27         {

28             if(i>=2)

29                 dfs(i-1,j);

30         }

31         map[i][j] /= 2;

32         if(map[i][j]%2 == 0) // 0 0 1 2 东边肯定没墙

33         {

34             if(j<=col-1)

35                 dfs(i,j+1);

36         }

37         map[i][j] /= 2;

38         if(map[i][j]%2 == 0) // 0 0 0 1 南边肯定没墙

39         {

40             if(i<=row-1)

41                 dfs(i+1,j);

42         }

43     }

44 }

45 int main()

46 {

47     while(cin>>row>>col)

48     {

49         memset(map,0,sizeof(map));

50         int i,j;        

51         for(i=1 ;i<=row ;i++)

52             for(j=1 ;j<=col ;j++)

53                 cin>>map[i][j];

54         ans = 0;

55         int count = 0;

56         memset(vis,false,sizeof(vis));

57         for(i=1 ;i<=row ;i++)

58             for(j=1 ;j<=col ;j++)

59             {

60                 if(!vis[i][j])

61                 {

62                      ans = 0;

63                      dfs(i,j);

64                      if(ans>area)

65                          area=ans;

66                     //联通的小方格算一个房间 

67                      if(ans)

68                          ++count;

69                  }

70              }

71         cout<<count<<endl;

72         cout<<area<<endl;

73     }

74     return 0;

75 }