ACM blockhouses
blockhouses
- 描述
- 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..
....
2
XX
.X
3
.X.
X.X
.X.
3
...
.XX
.XX
4
....
....
....
....
0 - 样例输出
-
5
1
5
2
4#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <iterator>
#include <utility>
using namespace std;
typedef pair<int,int> Point;
int ans = ; bool check(vector<string>& blockhouse,Point p){
for(int i = p.first; i>=; --i){
if(blockhouse[i][p.second] == 'X') break;
if(blockhouse[i][p.second] == '*') return false;
}
for(int i = p.second; i >= ; -- i){
if(blockhouse[p.first][i] == 'X') break;
if(blockhouse[p.first][i] == '*') return false;
}
return true;
} void dfs(vector<string>& blockhouse,Point pos,int sum ){
int n = blockhouse.size();
if(ans < sum) ans = sum;
if(pos.second == n) {pos.first++;pos.second =;}
if(pos.first == n && pos.second == ) return;
if(blockhouse[pos.first][pos.second] == 'X') dfs(blockhouse,Point(pos.first,pos.second+),sum);
else{
if(check(blockhouse, pos)) {
blockhouse[pos.first][pos.second] ='*';
dfs(blockhouse,Point(pos.first,pos.second+),sum+);
blockhouse[pos.first][pos.second] = '.';
}
dfs(blockhouse,Point(pos.first,pos.second+),sum);
}
} int main(){
int n;
while(cin >> n && n){
vector<string> blockhouse(n);
for(int i = ; i < n ; ++ i) cin >> blockhouse[i];
ans = ;
dfs(blockhouse,Point(,),);
cout<< ans<<endl; }
}
ACM blockhouses的更多相关文章
- SCNU ACM 2016新生赛决赛 解题报告
新生初赛题目.解题思路.参考代码一览 A. 拒绝虐狗 Problem Description CZJ 去排队打饭的时候看到前面有几对情侣秀恩爱,作为单身狗的 CZJ 表示很难受. 现在给出一个字符串代 ...
- SCNU ACM 2016新生赛初赛 解题报告
新生初赛题目.解题思路.参考代码一览 1001. 无聊的日常 Problem Description 两位小朋友小A和小B无聊时玩了个游戏,在限定时间内说出一排数字,那边说出的数大就赢,你的工作是帮他 ...
- acm结束了
最后一场比赛打完了.之前为了记录一些题目,开了这个博客,现在结束了acm,这个博客之后也不再更新了. 大家继续加油!
- 关于ACM的总结
看了不少大神的退役帖,今天终于要本弱装一波逼祭奠一下我关于ACM的回忆. 从大二上开始接触到大三下结束,接近两年的时间,对于大神们来说两年的确算不上时间,然而对于本弱来说就是大学的一半时光.大一的懵懂 ...
- 第一届山东省ACM——Phone Number(java)
Description We know that if a phone number A is another phone number B’s prefix, B is not able to be ...
- 第一届山东省ACM——Balloons(java)
Description Both Saya and Kudo like balloons. One day, they heard that in the central park, there wi ...
- ACM之鸡血篇
一匹黑马的诞生 故事还要从南京现场赛讲起,话说这次现场赛,各路ACM英雄豪杰齐聚南京,为争取亚洲总舵南京分舵舵主之职位,都使出了看 家本领,其中有最有实力的有京城两大帮清华帮,北大帮,南郡三大派上交派 ...
- 【codeforces 415D】Mashmokh and ACM(普通dp)
[codeforces 415D]Mashmokh and ACM 题意:美丽数列定义:对于数列中的每一个i都满足:arr[i+1]%arr[i]==0 输入n,k(1<=n,k<=200 ...
- acm 1002 算法设计
最近突然想往算法方向走走,做了做航电acm的几道题 二话不说,开始 航电acm 1002 题主要是处理长数据的问题,算法原理比较简单,就是用字符数组代替int,因为int太短需要处理的数据较长 下面是 ...
随机推荐
- 数据结构和算法 – 11.高级排序算法(下)
三.选择类排序 3.1.简单选择排序 http://www.cnblogs.com/tangge/p/5338734.html#XuanZe 3.2 堆排序 要知道堆排序,首先要了解一下二叉树的模型. ...
- 重温WCF之WCF中可靠性会话(十四)
1.WCF中可靠性会话在绑定层保证消息只会被传输一次,并且保证消息之间的顺序.当使用TCP(Transmission Control Protocol,传输控制协议)通信时,协议本身保证了可靠性.然而 ...
- Win10开机提示Resume from Hibernation该怎么办?
Windows10系统的电脑开机提示:Resume from Hibernation(从休眠恢复),这是电脑没有真正关机,而是上次关机时进入了[休眠状态],所以开机时提示:从休眠恢复.如何解决Wind ...
- 基于MyEclipse6.5的ssh整合
1.编写目的 为了学习,为了更好的学习java. 为了让想要学习这个整合的人少走弯路! ! ! 2.实验环境 l MyEclipse6.5 l JBoss4.2.1 l SQL2005 l 数据库脚本 ...
- 华为Mate8 NFC 时好时坏,怎么解决呢?
拿起手机朝桌子上磕几下,nfc就好用了.这是花粉总结的,我也试过,很灵.注意要带套,摄像头朝下,头部低一点往下磕.因为nfc芯片在头部,估计是接触不良.
- SQL常用方言列表
DB2 org.hibernate.dialect.DB2Dialect DB2 AS/400 org.hibernate.dialect.DB2400Dialect DB2 OS390 org.hi ...
- impdp导入job
结论: 10g to 10g:整个用户导出,无法正常导入JOB 10g to 11g:impdp时加SCHEMAS参数会导致无法正常导入JOB 11g to 11g:可以正常导入JOB 参见:http ...
- 阿里云(ECS)Centos服务器LNMP环境搭建
阿里云( ECS ) Centos7 服务器 LNMP 环境搭建 前言 第一次接触阿里云是大四的时候,当时在校外公司做兼职,关于智能家居项目的,话说当时俺就只有一个月左右的 php 后台开发经验(还是 ...
- JQuery常用函数及功能小结--转载
1.文档加载完成执行函数 $(document).ready(function(){ alert("开始了"); }); 2.添加/删除CSS类 $("#some-id ...
- Android Studio 插件整理
1.GsonFormat 快速将json字符串转换成一个Java Bean,免去我们根据json字符串手写对应Java Bean的过程. 使用方法:快捷键Alt+S也可以使用Alt+Insert选择G ...