2.1 castle

深度优先搜索

The Castle
IOI'94 - Day 1

In a stroke of luck almost beyond imagination, Farmer John was sent a ticket to the Irish Sweepstakes (really a lottery) for his birthday. This ticket turned out to have only the winning number for the lottery! Farmer John won a fabulous castle in the Irish countryside.

Bragging rights being what they are in Wisconsin, Farmer John wished to tell his cows all about the castle. He wanted to know how many rooms it has and how big the largest room was. In fact, he wants to take out a single wall to make an even bigger room.

Your task is to help Farmer John know the exact room count and sizes.

The castle floorplan is divided into M (wide) by N (1 <=M,N<=50) square modules. Each such module can have between zero and four walls. Castles always have walls on their "outer edges" to keep out the wind and rain.

Consider this annotated floorplan of a castle:

1   2   3   4   5   6   7

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

 1 #   |   #   |   #   |   |   #

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

 2 #   #   |   #   #   #   #   #

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

 3 #   |   |   #   #   #   #   #   

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

 4 # ->#   |   |   |   |   #   #   

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



#  = Wall     -,|  = No wall

-> = Points to the wall to remove to

     make the largest possible new room

By way of example, this castle sits on a 7 x 4 base. A "room" includes any set of connected "squares" in the floor plan. This floorplan contains five rooms (whose sizes are 9, 7, 3, 1, and 8 in no particular order).

Removing the wall marked by the arrow merges a pair of rooms to make the largest possible room that can be made by removing a single wall.

The castle always has at least two rooms and always has a wall that can be removed.

PROGRAM NAME: castle

INPUT FORMAT

The map is stored in the form of numbers, one number for each module, M numbers on each of N lines to describe the floorplan. The input order corresponds to the numbering in the example diagram above.

Each module number tells how many of the four walls exist and is the sum of up to four integers:

• 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.

Line 1:	Two space-separated integers: M and N
Line 2..:	M x N integers, several per line.

SAMPLE INPUT (file castle.in)

7 4

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

OUTPUT FORMAT

The output contains several lines:

Line 1:	The number of rooms the castle has.
Line 2:	The size of the largest room
Line 3:	The size of the largest room creatable by removing one wall
Line 4:	The single wall to remove to make the largest room possible

Choose the optimal wall to remove from the set of optimal walls by choosing the wall farthest to the west (and then, if still tied, farthest to the south). Name that wall by naming the module that borders it on either the west or south, along with a direction of N or E giving the location of the wall with respect to the module.

SAMPLE OUTPUT (file castle.out)

5

9

16

4 1 E



/*
ID: xdzhbin1
LANG: C++
TASK: castle
*/

#include <fstream>
#include <vector>
#include <iostream>
#include <queue>
#include <iostream>
using namespace std;

typedef enum
{
	WEST = 1,
	NORTH = 2,
	EAST = 4,
	SOUTH = 8
}Direction;

//向某个方向是否可以走通
bool across(Direction dir, int wall)
{
	return (15^wall)&dir;
}

//房间结构体
typedef struct room
{
	int wall;	//房间四面墙的情况
	int group;		//房间的分组
	int x;	//房间的所处的行
	int y;	//列
}room;

class Castle
{
private :
	int row,col; //城堡包含房间的行数和列数
public :
	int maxrooms; //最大连通房间的数目
	vector<int> v; //记录每组连通房间的数目
	room** rooms; //城堡包含的房间
	int pushx,pushy; //推倒墙的位置
	char pushDir;	//推倒墙的方向
	Castle(int r,int c):row(r),col(c)
	{
		int i;
		rooms = new room*[r];	//申请房间的内存空间
		for (i=0; i<r; i++)
		{
			rooms[i] = new room[c];
		}
	}
	~Castle()
	{
		int i;
		for (i=0; i<row; i++)	//释放
		{
			delete[] rooms[i]; 
		}
		delete[] rooms; 
	}
	int max()	//返回向量v中最大的元素
	{
		int i,max = v[0];
		for (i=1; i<v.size(); i++)
		{
			if (v[i]>max)
			{
				max = v[i];
			}
		}
		return max;
	}
	void search();	//给房间分组
	void push(); //推倒一面墙
};

void Castle::push()
{
	int i,j;
	//按规定顺序，尝试推倒每一面墙
	for (j=0; j<=col-1; j++)
	{
		for (i=row-1; i>=0; i--)
		{
			//尝试北面的墙
			if (i!=0 && rooms[i-1][j].group!=rooms[i][j].group
				&& v[rooms[i-1][j].group-1]+v[rooms[i][j].group-1]>maxrooms)
			{
				pushx = i;
				pushy = j;
				pushDir = 'N';
				maxrooms = v[rooms[i-1][j].group-1]+v[rooms[i][j].group-1];
			}
			//尝试东面的墙
			if (j!=col-1 && rooms[i][j].group!=rooms[i][j+1].group
				&& v[rooms[i][j].group-1]+v[rooms[i][j+1].group-1]>maxrooms)
			{
				pushx = i;
				pushy = j;
				pushDir = 'E';
				maxrooms = v[rooms[i][j].group-1]+v[rooms[i][j+1].group-1];
			}
		}
	}
}

//搜索最大连通房间的数目
//广度优先，运用队列
void Castle::search()
{
	int curGroup = 0;
	room curRoom;
	vector<int> groupNums;
	queue<room> queueRoom;
	int counter;
	
	int i,j;
	//遍历每个房间
	for (i=0; i<row; i++)
	{
		for (j=0; j<col; j++)
		{
			if (rooms[i][j].group==0) //新的分组
			{
				counter = 0;
				queueRoom.push(rooms[i][j]);
				curGroup++;
			}
			else //已经跟之前的某个房间连通
			{
				continue;
			}
			
			//标记所有跟当前房间连接的房间
			while (!queueRoom.empty())
			{
				curRoom = queueRoom.front();
				rooms[curRoom.x][curRoom.y].group = curGroup;
				queueRoom.pop();
				counter++;

				if (across(WEST,curRoom.wall)&&rooms[curRoom.x][curRoom.y-1].group==0)
				{
					rooms[curRoom.x][curRoom.y-1].group = curGroup;//改变入队列元素的组号，防止重复入队列
					queueRoom.push(rooms[curRoom.x][curRoom.y-1]);
				}
				if (across(NORTH,curRoom.wall)&&rooms[curRoom.x-1][curRoom.y].group==0)
				{
					rooms[curRoom.x-1][curRoom.y].group = curGroup;
					queueRoom.push(rooms[curRoom.x-1][curRoom.y]);
				}
				if (across(EAST,curRoom.wall)&&rooms[curRoom.x][curRoom.y+1].group==0)
				{
					rooms[curRoom.x][curRoom.y+1].group = curGroup;
					queueRoom.push(rooms[curRoom.x][curRoom.y+1]);
				}
				if (across(SOUTH,curRoom.wall)&&rooms[curRoom.x+1][curRoom.y].group==0)
				{
					rooms[curRoom.x+1][curRoom.y].group = curGroup;
					queueRoom.push(rooms[curRoom.x+1][curRoom.y]);
				}
			}
			v.push_back(counter);
		}
	}
	//maxrooms = *max_element(v.begin(),v.end());
	maxrooms = max();
}

int main()
{
	ifstream in("castle.in");
	ofstream out("castle.out");
	int r,c;
	in>>c>>r;
	Castle castle(r,c);
	int i,j;
	for (i=0; i<r; i++)
	{
		for (j=0; j<c; j++)
		{
			castle.rooms[i][j].x = i;
			castle.rooms[i][j].y = j;
			in>>castle.rooms[i][j].wall;
			castle.rooms[i][j].group = 0;
		}
	}
	castle.search();
	out<<castle.v.size()<<endl<<castle.maxrooms<<endl;
	castle.push();
	out<<castle.maxrooms<<endl;
	out<<castle.pushx+1<<" "<<castle.pushy+1<<" "<<castle.pushDir<<endl;
	
	
	/*for (i=0; i<r; i++)
	{
		for (j=0; j<c; j++)
		{
			cout<<castle.rooms[i][j].group<<" ";
		}
		cout<<endl;
	}
	for (i=0; i<castle.v.size(); i++)
	{
		cout<<castle.v[i]<<" ";
	}*/
	return 0;
}