POJ 1164 The Castle （深搜）

The Castle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8241		Accepted: 4648

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

大概题意：

就是给定一个m*n的矩阵，有很多的空间，有的是由墙隔开的，现在要找到一共有多少个房间，以及最大的房间是多大？并把它两打印出来。
具体思路：
第一：定义一个存储矩阵数据的二维数组，以及用来做标记的二维数组
第二：西是1，北是2，东是4，南是8，因此要判断是否能通过，只要对date中的数据依次和1,2,4,8相与，就知道了是否能过通行。
第三：就是一个深搜，也就是我们所说的完美的递归调用。

具体代码实现如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int r,c,roomarea,roomnum = 0,maxroom = 0;
int room[60][60],visit[60][60];
void dfs(int i,int j)
{
    if(visit[i][j])
        return;
    roomarea++;
    visit[i][j] = roomnum;
    if(!(room[i][j] & 1))
        dfs(i,j - 1);
    if(!(room[i][j] & 2))
        dfs(i - 1,j);
    if(!(room[i][j] & 4))
        dfs(i,j + 1);
    if(!(room[i][j] & 8))
        dfs(i + 1,j);
}
int main()
{
    scanf("%d %d",&r,&c);
    for(int i = 1;i <= r; i++)
    {
        for(int j = 1;j <= c; j++)
        {
            scanf("%d",&room[i][j]);
        }
    }
    memset(visit,0,sizeof(visit));
    for(int i = 1;i <= r; i++)
    {
        for(int j = 1;j <= c; j++)
        {
            if(!visit[i][j])
            {
                roomnum++;
                roomarea = 0;
                dfs(i,j);
                maxroom = max(maxroom,roomarea);
            }
        }
    }
    cout << roomnum << endl << maxroom;
    return 0;
}