回溯搜索166:The Castle(OpenJudge…

166:The Castle

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时间限制: 

1000ms

 

内存限制: 

65536kB

描述

     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 thatcalculates
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 havebetween zero and four walls. 

输入

Your program is to read from standard input. The first linecontains the number of modules in the north-south direction and thenumber of modules in the east-west direction. In the followinglines each module is described by a number (0 <= p<= 15). This number is the sum of: 1 (= wall to thewest), 2 (= wall to the north), 4 (= wall to the east), 8 (= wallto the south). Inner walls are defined twice; a wall to the southin module 1,1 is also indicated as a wall to the north in module2,1. The castle always has at least two rooms.

输出

Your program is to write to standard output: First the number ofrooms, then the area of the largest room (counted in modules).

样例输入

4711 6 11 6 3 10 67 9 6 13 5 15 51 10 12 7 13 7 513 11 10 8 10 12 13

样例输出

59

源码（OpenJudge提交通过）

#include"stdio.h"

#include"stdlib.h"

struct module

{

char north;

char south;

char east;

char west; //上下左右四面墙 

int sum;

int sr;

int sc; // (sr,sc)记录到达此位置的上一个位置，以便返回 

bool flag;

};

int main()

{

int i,j,m , n; //m行，n列

scanf("%d",&m);

scanf("%d",&n);

struct module md[50][50];

//struct module **md=(struct module**)malloc(sizeof(structmodule)*m); 

for(i=0;i<m;i++)

{

/

if(md[i][j].sum%2==1)

md[i][j].west='y';

else

md[i][j].west='n';

md[i][j].sum/=2;

if(md[i][j].sum%2==1)

md[i][j].north='y';

else

md[i][j].north='n';

md[i][j].sum/=2;

if(md[i][j].sum%2==1)

md[i][j].east='y';

else

md[i][j].east='n';

md[i][j].sum/=2;

if(md[i][j].sum%2==1)

md[i][j].south='y';

else

md[i][j].south='n';

md[i][j].flag=false;

}

}

int r , c; //记录当前搜索的行和列 

int NumRoom ,MaxRoom,temp,count;

count=0; //标记被占有的module 

NumRoom=0;

MaxRoom=0;

bool fg;

while(count<m*n)

{

fg=false;

for(r=0;r<m;r++)

{

for(c=0;c<n;c++)

{ if(!md[r][c].flag)

{

fg=true;

break;

}

}

if(fg)

break;

}

md[r][c].flag=true;

count++;

temp=1;

i=r;

j=c; //记录开始搜索的位置

md[r][c].sr=r;

md[r][c].sc=c;

int si,sj;

int redo=0;//

do{

if(md[r][c].west=='n' &&c>0 &&!md[r][c-1].flag)

{

md[r][c-1].sr=r;

md[r][c-1].sc=c;

c--;

count++;

temp++; 

md[r][c].flag=true;

}//左走，west

else if(md[r][c].north=='n' &&r>0 &&!md[r-1][c].flag)

{

md[r-1][c].sr=r;

md[r-1][c].sc=c;

r--;

count++;

temp++; 

md[r][c].flag=true;

}//上走，north

else if(md[r][c].east=='n' &&c<n-1 &&!md[r][c+1].flag)

{

md[r][c+1].sr=r;

md[r][c+1].sc=c;

c++;

count++;

temp++; 

md[r][c].flag=true;

}//右走，east

else if(md[r][c].south=='n' &&r<n-1 &&!md[r+1][c].flag)

{

md[r+1][c].sr=r;

md[r+1][c].sc=c;

r++;

count++;

temp++; 

md[r][c].flag=true;

}//下走，south

else{

si=md[r][c].sr;

sj=md[r][c].sc;

r=si;

c=sj;

}//返回上一个位置 

if(r==i && c==j)

redo++; //在每个开始搜索的位置重复4次

}while(redo<4 );

NumRoom++;

if(temp>MaxRoom)

MaxRoom=temp;

}

printf("%d\n%d\n",NumRoom,MaxRoom);

return 0;

}

运行结果