Problem
DescriptionSuppose 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 4×4) 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`
分析：
如题目上的图片所示，圆代表的是枪，方格代表的是墙。在同一行或者同一列的枪之间可以相互射击，因此枪不可以放在同一行或者同一列，但是如果他们之间有墙阻隔的话，就认为是不能够被射击到的，这样的话就可以放在同一行或者同一列。
类似于n皇后的问题，我们从第一个点开始找，每次在找的时候只看它所在行的左边和所在列的上边，看能不能够放置枪，能的话就把枪的数目加1，再往后找，不能的话直接往后找。
代码：
#include<iostream>
#include<stdio.h>
#include<queue>
#include<algorithm>
#include<map>
#include<string.h>
using namespace std;
int n,Min=0;
char tu[5][5];
bool ZhangAi(int x,int y)//判断(x,y)这个点能不能够放枪
{
for(int i=x-1;i>=0;i--)//在该列的上方寻找
{
if(tu[i][y]=='*')//如果该列放过枪了，就肯定不能够放了
return false;
if(tu[i][y]=='X')//如果该列有墙，也就意味着能够放，不用再往上寻找了
break;
}
for(int i=y-1;i>=0;i--)
{
if(tu[x][i]=='*')//如果该行放过枪了，就肯定不能够放了
return false;
if(tu[x][i]=='X')//如果该行有墙，也就意味着能够放，不用再往左寻找了
break;
}
return true;//返回true是该行的左边，列的上边，都没有放过枪或者遇到了墙
} void dfs(int cur,int s)
{
if(cur==n*n)//当前的放个访问的第cur个放个，
{
if(s>Min)
Min=s;
return;
}
//当前的放个访问的第cur个方格，除以列数就是所在的行，对列数取余，就是所在的列
int x=cur/n;
int y=cur%n;
if(tu[x][y]=='.'&&ZhangAi(x,y))//判断这个点能不能放枪
{
tu[x][y]='*';
dfs(cur+1,s+1);
tu[x][y]='.';//因为递归调用，要标记后再把标记释放掉
}
dfs(cur+1,s);
}
int main()
{
while(~scanf("%d",&n)&&n)
{
Min=0;
memset(tu,'.',sizeof(tu));
for(int i=0;i<n;i++)
{
scanf(" %s",tu[i]);
}
dfs(0,0);
printf("%d\n",Min);
}
return 0;
}
