Fire Net
A blockhouse is a small castle that has four openings through which to shoot. The four openings are facing North, East, South, and West, respectively. There will be one machine gun shooting through each opening.
Here we assume that a bullet is so powerful that it can run across any distance and destroy a blockhouse on its way. On the other hand, a wall is so strongly built that can stop the bullets.
The goal is to place as many blockhouses in a city as possible so that no two can destroy each other. A configuration of blockhouses is legal provided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them. In this problem we will consider small square cities (at most 4x4) that contain walls through which bullets cannot run through.
The following image shows five pictures of the same board. The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways.
Your task is to write a program that, given a description of a map, calculates the maximum number of blockhouses that can be placed in the city in a legal configuration.
Input
Output
Sample Input
Sample Output
#include <iostream>
#include <cstdio> using namespace std; char str[][];
int dir[][]={{,},{,-},{,},{-,}};// up down left right 四个方向
int ans,n;
int judge(int r,int c)// 判断当前位置能否放炮台
{
int i;
bool flag=;//默认当前位置符合条件
for(i=;i<;i++)//对当前位置,跑完它的四个方位
{
int rr,cc;
rr=r;
cc=c;
while()
{
rr=rr+dir[i][];
cc=cc+dir[i][];
if(rr< || cc< || rr>n || cc>n)//当到达边界时跳出
break;
else if( str[rr][cc] == '@')
{
flag=;
break;
}//当前位置所在行,列已有炮台,不符合条件
else
{
if(str[cc][rr] == 'X')
break;
}//遇到碉堡,跳出
}//当前位置是否符合要求
}
return flag;
}
void dfs(int res)
{
for(int i= ; i<=n ; i++)
{
for(int j= ; j<=n ; j++)
{
if(str[i][j]=='.' && judge(i,j))
{
str[i][j]='@';//当前位置是空地,且可以放炮台
dfs(res+);//当前位置符合要求,进行下一层搜索
str[i][j]='.';//记得还原
}
}
}
if(res>ans)
ans=res;//当前最优解
return;
}
int main()
{
int i,j;
while(~scanf("%d%*c",&n) && n)
{
ans=;
for(i= ; i<=n ; i++)
{
for(j= ; j<=n ; j++)
cin>>str[i][j];
}
dfs();
cout<<ans<<endl;
}
return ;
}