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[usaco] 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 module farthest to the west (and then, if still tied, farthest to the south). If still tied, choose 'N' before 'E'. 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
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本題的關鍵點
1：如果保存點和邊的信息。
把每個小的格子看成一個點，如果兩個格子同屬於一個房間的話，表明兩個點之間存在一個邊。
由於題目中已經給出了明確的信息：每個格子四周的牆是用二進製數來表示的。因此，隻需要M*N個點的信息，
每個點要儲存：點是否被訪問過，點所屬的連通分量，點四周的牆信息。
2：如何查找room的數量和最大的room的麵積。
利用BFS，遍曆所有的節點，把每個節點所屬的component進行初始化。
3：尋找可能連接兩個房間構成一個最大房間的牆
遍曆每一個牆，找出最大的。
我的代碼：
--------------------------------------------------------------------------------------------------
/*
ID: yunleis2
PROG: castle
LANG: C++
*/
#include<iostream>
#include<fstream>
#include<queue>
#include<vector>
using namespace std;
typedef unsigned int uint;
class vertex
{
public:
uint wall;
int component;
int visited;
vertex()
{
component=-1;
visited=0;
}
};
static uint WALL=1;
static uint CROSS=2;
int main()
{
fstream fin("castle.in",ios::in);
int M,N;
fin>>M>>N;
vertex * src=new vertex[M*N];
vector<int > rooms;
for(int i=0;i<N;i++)
{
for(int j=0;j<M;j++)
{
fin>>src[i*M+j].wall;
}
}
edge * lr1=new edge[(M-1)*N];
edge * hl2=new edge [(N-1)*M];
for(int i=0;i<N;i++)
{
for(int j=0;j<(M-1);j++)
{
uint wall=src[i*M+j].wall;
if((wall&4)!=0)
lr1[i*(M-1)+j].type=WALL;
else
{ lr1[i*(M-1)+j].type=CROSS;
lr1[i*(M-1)+j].left=j;
lr1[i*(M-1)+j].right=j+1;
}
}
}
int g_cmpnt=1;
int g_large=0;
for(int i=0;i<(M*N);i++)
{
if(src[i].visited==0)
{
/************************************************************************/
/* bfs */
/************************************************************************/
queue<int> q;
q.push(i);
int t_large=0;
int t_cmpnt=g_cmpnt++;
src[i].component=t_cmpnt;
while(!q.empty())
{
int tmp=q.front();
q.pop();
if(src[tmp].visited==0)
{
src[tmp].component=t_cmpnt;
src[tmp].visited=1;
t_large++;
if((src[tmp].wall&1)==0)
{
q.push(tmp-1);
}
if((src[tmp].wall&2)==0)
{
q.push(tmp-M);
}
if((src[tmp].wall&4)==0)
{
q.push(tmp+1);
}
if((src[tmp].wall&8)==0)
{
q.push(tmp+M);
}
}
}
if(t_large>g_large)
g_large=t_large;
rooms.push_back(t_large);
}
}
fstream fout("castle.out",ios::out);
fout<<g_cmpnt-1<<endl;
fout<<g_large<<endl;
//search all the walls;
int more_large=g_large;
int r,c;
char direc;
for(int i=0;i<M;i++)
{
for(int j=N-1;j>0;j--)
{
if((src[j*M+i].wall&2)!=0)
{
if(src[j*M+i].component!=src[(j-1)*M+i].component)
{
int t1;
if((t1=rooms.at(src[j*M+i].component-1)+rooms.at(src[(j-1)*M+i].component-1))>more_large)
{
more_large=t1;
r=j+1;c=i+1;
direc='N';
}
}
}
}
if(i==(M-1))
continue;
for(int j=N-1;j>=0;j--)
{
if((src[j*M+i].wall&4)!=0)
{
if(src[j*M+i].component!=src[(j)*M+i+1].component)
{
int t1;
if((t1=rooms.at(src[j*M+i].component-1)+rooms.at(src[(j)*M+i+1].component-1))>more_large)
{
more_large=t1;
r=j+1;c=i+1;
direc='E';
}
}
}
}
}
fout<<more_large<<endl;
fout<<r<<" "<<c<<" "<<direc<<endl;
}
最後更新：2017-04-02 06:51:49