POJ 1164：The Castle

The Castle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6677		Accepted: 3767

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

题意是一个城堡分成了m*n个块，然后给出了每个块一个数字，这个数字代表门的情况，如果这个块西面有门，那么1就要加到这个数字中。如果这个块北面有门，那么2就要加到这个数字中。如果这个块东面有门，那么4就要加到这个数字中。如果这个块南面有门，那么8就加到这个数字中。

所以就可以使用这个数&1 &2 &4 &8来判断哪一个方向有门，连通着的算一个房间，要求的是房间的数量和最大房间的块数。

应该算是深度搜索的模板题了吧。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int row,col,i,j,sum,result1,result2;
int value[70][70];
int color[70][70];

int move_x[5]={1,0,-1,0};
int move_y[5]={0,1,0,-1};

void dfs(int a,int b)
{
	color[a][b]=1;
	sum++;
	int a_x,b_x;

	if((value[a][b]&1)==0)
	{
		a_x=a+move_x[3];
		b_x=b+move_y[3];
		if(a>=1 && a<=row && b>=1 && b<=col && color[a_x][b_x]==0)
		{
			dfs(a_x,b_x);
		}
	}
	if((value[a][b]&2)==0)
	{
		a_x=a+move_x[2];
		b_x=b+move_y[2];
		if(a>=1 && a<=row && b>=1 && b<=col && color[a_x][b_x]==0)
		{
			dfs(a_x,b_x);
		}
	}
	if((value[a][b]&4)==0)
	{
		a_x=a+move_x[1];
		b_x=b+move_y[1];
		if(a>=1 && a<=row && b>=1 && b<=col && color[a_x][b_x]==0)
		{
			dfs(a_x,b_x);
		}
	}
	if((value[a][b]&8)==0)
	{
		a_x=a+move_x[0];
		b_x=b+move_y[0];
		if(a>=1 && a<=row && b>=1 && b<=col && color[a_x][b_x]==0)
		{
			dfs(a_x,b_x);
		}
	}

}

int main()
{
	memset(color,0,sizeof(color));

	cin>>row>>col;
	for(i=1;i<=row;i++)
	{
		for(j=1;j<=col;j++)
		{
			cin>>value[i][j];
		}
	}
	result1=0;
	result2=0;
	for(i=1;i<=row;i++)
	{
		for(j=1;j<=col;j++)
		{
			sum=0;
			if(color[i][j]==0)
			{
				dfs(i,j);
				result2++;
			}
			result1=max(sum,result1);
		}
	}

	cout<<result2<<endl;
	cout<<result1<<endl;
	return 0;
}