杭电1045（zoj1002）Fire Net

 Fire Net
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7905    Accepted Submission(s): 4499

Problem Description


Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall.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


The input file contains one or more map descriptions, followed by a line containing the number 0 that signals the end of the file. Each map description begins with a line containing a positive integer n that is the size of the city; n will be at most 4. The next n lines each describe one row of the map, with a '.' indicating an open space and an uppercase 'X' indicating a wall. There are no spaces in the input file.


 


Output


For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.


 


Sample Input




4 .X.. .... XX.. .... 2 XX .X 3 .X. X.X .X. 3 ... .XX .XX 4 .... .... .... .... 0



 




Sample Output




5 1 5 2 4



 就是说黑点是机关枪，黑块是墙，墙能够挡子弹，机关枪特别厉害，能够穿透其所在的行和列上的所有东西（除了黑块），也就是说每一行每一列只能有一把枪，或者有个墙，问的是计算出合理分配机关枪的最大数目。


思路是判断判断该单元格的行和列的合法性，如果合法，就能放置一个机枪。


用深搜.


#include
#include
#include
using namespace std;
char cmap[5][5];//设置规格 
int zy;
int n,i,j,k,l; 
bool cp(int x,int y)//判断 该点是否合法 
{
	int i;
	for(i=x-1;i>=0;i--)
	{
		if(cmap[i][y]=='0')//判断该点的行 是否合法 
		return false;
		if(cmap[i][y]=='X')//如果左边有墙而不是枪，就一定合法 
		break;
	}
	for(i=y-1;i>=0;i--)
	{
		if(cmap[x][i]=='0')//判断该点的了是否合法 
		return false;
		if(cmap[x][i]=='X')//如果下边有墙而不是枪，就一定合法 
		break;
	}
	return true;
}
void dfs(int k,int js)//深入搜索 
{
	int x,y;
	if(k==n*n)//即搜索完成 
	{
		if(js>zy)//更新最优解 
		{
			zy=js;
			return ;
		}
	}
	else
	{
		x=k/n;//行的坐标 
		y=k%n;//列的坐标 
		if(cmap[x][y]=='.'&&cp(x,y))//如果这一点不是墙还合法 
		{
			cmap[x][y]='0';//标记这一点 用过了 
			dfs(k+1,js+1);//枪数+1； 
			cmap[x][y]='.';
		}
	 dfs(k+1,js);//如果这一点是墙或者不合法	，搜索下一个点，枪数不变 
	}
}
int main()
{
	while(scanf("%d",&n),n)
	{
		zy=0;
		for(i=0;i