hdu1045 Fire Net (dfs)_hdu 1045-程序员宅基地

Fire Net

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11697 Accepted Submission(s): 7014

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

Source

Zhejiang University Local Contest 2001

Recommend

We have carefully selected several similar problems for you: 1067 1258 1053 1789 1026

Statistic | Submit | Discuss | Note

由于这道题的n最大才为4 所以可以深搜。其实我的第一印象是二分匹配的

没办法喜欢搜索

#include <stdio.h>
#include <string.h>
int n,sum;
char map[10][10];
bool vis[10][10];
//边界检测 
bool limit(int x,int y)
{
	if(x<0||y<0||x>=n||y>=n) return false;
	return true;
}
//检测 
bool check(int x,int y)
{
	//检查第x行
	for(int i=y-1;i>=0;i--)
	{
		if(vis[x][i]) return false;
		if(map[x][i]=='X') break;
	}
	for(int i=y+1;i<n;i++)
	{
		if(vis[x][i]) return false;
		if(map[x][i]=='X') break;
	}
	//检查第y列 
	for(int i=x-1;i>=0;i--)
	if(vis[i][y]||map[i][y]=='X') 
	{
		if(vis[i][y]) return false;
		if(map[i][y]=='X') break;
	}
	for(int i=x+1;i<n;i++)
	{
		if(vis[i][y]) return false;
		if(map[i][y]=='X') break;
	}
	return true;
}
void dfs(int flag,int count)
{
	int x=flag/n,y=flag%n;

//	printf("-%d- ",count);
	if(count>sum) sum=count;
	if(!limit(x,y)) return ;
	
	if(map[x][y]!='X'&&!vis[x][y]&&check(x,y)) 
	{	
		vis[x][y]=true;
	//	printf("%d %d \n",x,y);
		dfs(flag+1,count+1);
	//	printf("\n*******************************\n");
		vis[x][y]=false;
	}
	dfs(flag+1,count);
}
int main()
{
	//freopen("in.txt","r",stdin);
	while(~scanf("%d",&n)&&n)
	{
		memset(vis,false,sizeof(vis));
		memset(map,0,sizeof(map));
		for(int i=0;i<n;i++)
		scanf("%s",map[i]);
		sum=0;
		dfs(0,0);
		printf("%d\n",sum);
	}
	return 0;
}

本文链接：https://blog.csdn.net/su20145104009/article/details/70213486

原作者删帖不实内容删帖广告或垃圾文章投诉

智能推荐

c# 调用c++ lib静态库_c#调用lib-程序员宅基地

文章浏览阅读2w次，点赞7次，收藏51次。四个步骤1.创建C++ Win32项目动态库dll 2.在Win32项目动态库中添加外部依赖项 lib头文件和lib库3.导出C接口4.c#调用c++动态库开始你的表演...①创建一个空白的解决方案，在解决方案中添加 Visual C++ , Win32 项目空白解决方案的创建：添加Visual C++ , Win32 项目这......_c#调用lib

deepin/ubuntu安装苹方字体-程序员宅基地

文章浏览阅读4.6k次。苹方字体是苹果系统上的黑体，挺好看的。注重颜值的网站都会使用，例如知乎：font-family: -apple-system, BlinkMacSystemFont, Helvetica Neue, PingFang SC, Microsoft YaHei, Source Han Sans SC, Noto Sans CJK SC, W..._ubuntu pingfang

html表单常见操作汇总_html表单的处理程序有那些-程序员宅基地

文章浏览阅读159次。表单表单概述表单标签表单域按钮控件demo表单标签表单标签基本语法结构<form action="处理数据程序的url地址“ method=”get|post“ name="表单名称”></form><!--method将表单中的数据传送给服务器处理，get方式直接显示在url地址中，数据可以被缓存，且长度有限制；而post方式数据隐藏传输，_html表单的处理程序有那些

PHP设置谷歌验证器（Google Authenticator）实现操作二步验证_php otp 验证器-程序员宅基地

文章浏览阅读1.2k次。使用说明:开启Google的登陆二步验证（即Google Authenticator服务）后用户登陆时需要输入额外由手机客户端生成的一次性密码。实现Google Authenticator功能需要服务器端和客户端的支持。服务器端负责密钥的生成、验证一次性密码是否正确。客户端记录密钥后生成一次性密码。下载谷歌验证类库文件放到项目合适位置(我这边放在项目Vender下面)https://github.com/PHPGangsta/GoogleAuthenticatorPHP代码示例://引入谷_php otp 验证器

【Python】matplotlib.plot画图横坐标混乱及间隔处理_matplotlib更改横轴间距-程序员宅基地

文章浏览阅读4.3k次，点赞5次，收藏11次。matplotlib.plot画图横坐标混乱及间隔处理_matplotlib更改横轴间距

docker — 容器存储_docker 保存容器-程序员宅基地

文章浏览阅读2.2k次。①Storage driver 处理各镜像层及容器层的处理细节，实现了多层数据的堆叠，为用户提供了多层数据合并后的统一视图②所有 Storage driver 都使用可堆叠图像层和写时复制（CoW）策略③docker info 命令可查看当系统上的 storage driver主要用于测试目的，不建议用于生成环境。_docker 保存容器

随便推点

网络拓扑结构_网络拓扑csdn-程序员宅基地

文章浏览阅读834次，点赞27次，收藏13次。网络拓扑结构是指计算机网络中各组件（如计算机、服务器、打印机、路由器、交换机等设备）及其连接线路在物理布局或逻辑构型上的排列形式。这种布局不仅描述了设备间的实际物理连接方式，也决定了数据在网络中流动的路径和方式。不同的网络拓扑结构影响着网络的性能、可靠性、可扩展性及管理维护的难易程度。_网络拓扑csdn

JS重写Date函数，兼容IOS系统_date.prototype 将所有 ios-程序员宅基地

文章浏览阅读1.8k次，点赞5次，收藏8次。IOS系统Date的坑要创建一个指定时间的new Date对象时，通常的做法是：new Date("2020-09-21 11:11:00")这行代码在 PC 端和安卓端都是正常的，而在 iOS 端则会提示 Invalid Date 无效日期。在IOS年月日中间的横岗许换成斜杠，也就是new Date("2020/09/21 11:11:00")通常为了兼容IOS的这个坑，需要做一些额外的特殊处理，笔者在开发的时候经常会忘了兼容IOS系统。所以就想试着重写Date函数，一劳永逸，避免每次ne_date.prototype 将所有 ios