hdu 1045 fire net &nyoj 587 blockhouses（dfs！）

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. 

输入

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.

输出

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

样例输入

4.X......XX......2XX.X3.X.X.X.X.3....XX.XX4................0

样例输出

51524

import java.util.Scanner;public class Main {static int max,n;static char map[][];public static boolean search(int x,int y){//如果在同行同列就不可以放置for(int i=x-1;i>=0;i--){if(map[i][y]=='P')return false;if(map[i][y]=='X')break;}for(int j=y-1;j>=0;j--){if(map[x][j]=='P')return false;if(map[x][j]=='X')break;}return true;}public static void dfs(int id,int num){//i代表已经搜索到的点，num代表该次搜索可以放得最多blockhouse//搜索完所有可能情况，每点'.'都会作为起点搜索一遍if(id==n*n){if(num>max)max=num;return;}////从左到右，从上到下依次给每个点编号，1、2、...、//求出当前点所在的行和列int row=id/n;int col=id%n;//搜索完所有可能情况，每点'.'都会作为起点搜索一遍if(map[row][col]=='.'&&search(row,col)){map[row][col]='P';dfs(id+1,num+1);map[row][col]='.';}dfs(id+1, num);}public static void main(String[] args) { Scanner scanner=new Scanner(System.in); while(true) { n=scanner.nextInt(); if(n==0)break; String save[]=new String[n]; map=new char[n+1][n+1]; for(int i=0;i<n;i++) {save[i]=scanner.next();for(int j=0;j<n;j++){map[i][j]=save[i].charAt(j);} } max=0; dfs(0,0); System.out.println(max); }}}