Fire Net HDU
Fire Net
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 13 Accepted Submission(s) : 5
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
Problem Description
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
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
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#include<Iostream>//代码写的难看了点..........可再写的简便一点
using namespace std;
void bianli(int a[5][5],int i,int j)
{
int ss;
for(ss=i-1;ss>=1;ss--)
if(a[ss][j]!=50)
a[ss][j]=20;
else
break;
for(ss=i+1;ss<=4;ss++)
if(a[ss][j]!=50)
a[ss][j]=20;
else
break;
for(ss=j-1;ss>=1;ss--)
if(a[i][ss]!=50)
a[i][ss]=20;
else
break;
for(ss=j+1;ss<=4;ss++)
if(a[i][ss]!=50)
a[i][ss]=20;
else
break;
}
int main()
{
int a[5][5],i,j,k,n,m,ans,ss;
char s;
while(cin>>k&&k)
{
ans=0;
for(i=1;i<=k;i++)
for(j=1;j<=k;j++)
{cin>>s;
if(s=='.')
a[i][j]=0;
else
a[i][j]=50;
}
for(i=1;i<=k;i++)
for(j=1;j<=k;j++)
{
if(a[i][j]!=50)
{
for(ss=i-1;ss>=1;ss--)
if(a[ss][j]!=50)
a[i][j]++;
else
break;
for(ss=i+1;ss<=4;ss++)
if(a[ss][j]!=50)
a[i][j]++;
else
break;
for(ss=j-1;ss>=1;ss--)
if(a[i][ss]!=50)
a[i][j]++;
else
break;
for(ss=j+1;ss<=4;ss++)
if(a[i][ss]!=50)
a[i][j]++;
else
break;
} }
/*for(i=1;i<=k;i++)
{
for(j=1;j<=k;j++)
cout<<a[i][j]<<' ';
cout<<endl;}*/
for(n=0;n<=9;n++)
for(i=1;i<=k;i++)
for(j=1;j<=k;j++)
{if(a[i][j]==n)
{ bianli(a,i,j);
ans++;}}
cout<<ans<<endl; }
return 0;
}
Fire Net HDU的更多相关文章
- (匹配)Fire Net --hdu --1045
链接: http://acm.hdu.edu.cn/showproblem.php?pid=1045 http://acm.hust.edu.cn/vjudge/contest/view.action ...
- Fire Net HDU 1045
简单深搜,可以完全暴力,不会超时的. #include<iostream> #include<cstring> #include<cmath> using name ...
- A - Fire Net - hdu 1045(二分图匹配)
题意:一个阵地可以向四周扫射,求出来最多能修多少个阵地,墙不可以被扫射透,阵地不能同行或者或者列(有墙隔着例外) 分析:很久以前就做过这道题..当时是练习深搜来着,不过时间复杂度比较高,现在再看突然发 ...
- Fire Net HDU - 1045(二分匹配)
把每一列中相邻的 . 缩为一个点 作为二分图的左边 把每一行中相邻的 . 缩为一个点 作为二分图的右边 然后求最大匹配即可 这题用匈牙利足够了...然而..我用了hk...有点大材小用的感觉// ...
- Fire Net HDU - 1045 (二分图匹配)
题意: 给出一张图,图中'X'表示wall,'.'表示空地,可以放置blockhouse同一条直线上只能有一个blockhouse,除非有wall 隔开,问在给出的图中最多能放置多少个blockhou ...
- KUANGBIN带你飞
KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题 //201 ...
- [kuangbin带你飞]专题1-23题目清单总结
[kuangbin带你飞]专题1-23 专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 Fli ...
- ACM--[kuangbin带你飞]--专题1-23
专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find T ...
- hdu 1045 Fire Net(最小覆盖点+构图(缩点))
http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit:1000MS Memory Limit:32768KB ...
随机推荐
- Android 开源项目android-open-project工具库解析之(一) 依赖注入,图片缓存,网络相关,数据库orm工具包,Android公共库
一.依赖注入DI 通过依赖注入降低View.服务.资源简化初始化.事件绑定等反复繁琐工作 AndroidAnnotations(Code Diet) android高速开发框架 项目地址:https: ...
- android doGet和doPost
doGet和doPost的差别 get和post是http协议的两种方法,另外还有head, delete等 这两种方法有本质的差别,get仅仅有一个流,參数附加在url后.大小个数有严格限制且仅仅 ...
- JAVA进阶----ThreadPoolExecutor机制(转)
ThreadPoolExecutor机制 一.概述 1.ThreadPoolExecutor作为java.util.concurrent包对外提供基础实现,以内部线程池的形式对外提供管理任务执行,线程 ...
- DOM - Element 对象
http://www.runoob.com/dom/dom-element.html Element 对象 Element 对象代表 XML 文档中的一个元素.元素可以包含属性.其他元素或文本.如果一 ...
- cocos2dx游戏开发学习笔记3-lua面向对象分析
在lua中,能够通过元表来实现类.对象.继承等.与元表相关的方法有setmetatable().__index.getmetatable().__newindex. 详细什么是元表在这里就不细说了,网 ...
- mysql 数据库备份ubuntu
安装 1 sudo apt-get update 2. sudo apt-get install mysql-server 3 sudo apt-get install mysql-client 4 ...
- javascript面向对象程序设计
在学习js面向对象编程之前,首先须要知道什么是面向对象.面向对象语言都有类的概念,通过它能够创建具有同样属性和方法的对象.但js并没有类的概念,因此js中的对象和其它语言的对象有所不同. js对象能够 ...
- [置顶] android系统功能调用大全
1.从google搜索内容 Intent intent = new Intent(); intent.setAction(Intent.ACTION_WEB_SEARCH); intent.pu ...
- STL之Vector(不定长数组)
vector是同一种对象的集合,每一个对象都有一个相应的整数索引值.和string对象一样,标准库将负责管理与存储元素相关的类存. 引入头文件 #include<vector> 1.vec ...
- Spring相框
1.什么是Spring相框?Spring有哪些主要模块框架? Spring框架是一个为Java应用程序的开发提供了综合.广泛的基础性支持的Java平台. Spring帮助开发人员攻克了开发中基础性的问 ...