Fire Net--hdu1045
Fire Net
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7929 Accepted Submission(s):
4521
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.
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.
maximum number of blockhouses that can be placed in the city in a legal
configuration.
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char map[][];
int n,now,ibest,current;
bool canput(int row,int col)//判断能不能放
{
int i;
for(i=row-;i>=;i--)//判断行
{
if(map[i][col]=='O')
return false;
if(map[i][col]=='X')
break;
}
for(i=col-;i>=;i--)//判断列
{
if(map[row][i]=='O')
return false;
if(map[row][i]=='X')
break;
}
return true;
}
void dfs(int k,int current)
{
int x,y;
if(k==n*n)//如果遍历完就返回
{
if(current>ibest)//更新最大的个数
ibest=current;
return ;
}
else
{
x=k/n;
y=k%n;
if(map[x][y]=='.'&&canput(x,y))
{
map[x][y]='O';
dfs(k+,current+);//下一次递归
map[x][y]='.';
}
dfs(k+,current);//当前不放碉堡
}
}
int main()
{
while(scanf("%d",&n),n)
{
getchar();
int k=,i;
ibest=;
current=;
for(i=;i<n;i++)
scanf("%s",map[i]);
dfs(,);
printf("%d\n",ibest);
}
return ;
}
Fire Net--hdu1045的更多相关文章
- HDU1045 Fire Net(DFS)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit: 2000/1000 MS (Java/Others) ...
- HDU-1045 Fire Net
http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit: 2000/1000 MS (Java/Others) Me ...
- HDU1045 Fire Net(DFS枚举||二分图匹配) 2016-07-24 13:23 99人阅读 评论(0) 收藏
Fire Net Problem Description Suppose that we have a square city with straight streets. A map of a ci ...
- HDU1045:Fire Net(二分图匹配 / DFS)
Fire Net Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- HDU1045 Fire Net —— 二分图最大匹配
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1045 Fire Net Time Limit: 2000/1000 MS (Java/Others) ...
- hdu-1045.fire net(缩点 + 二分匹配)
Fire Net Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- hdu1045 Fire Net
在一张地图上建立碉堡(X),要求每行没列不能放两个,除非中间有强挡着.求最多能放多少个碉堡 #include<iostream> #include<cstdio> #inclu ...
- hdu1045 Fire Net---二进制枚举子集
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1045 题目大意: 给你一幅n*n的图,再给你一些点,这些点的上下左右不能再放其他点,除非有墙('X') ...
- 【HDU-1045,Fire Net-纯暴力简单DFS】
原题链接:点击! 大致题意:白块表示可以放置炮台的位置——每个炮台可以攻击到上下左右的直线上的炮台(也就是说在它的上下左右直线上不可以再放置炮台,避免引起互相攻击),黑块表示隔离墙的位置——不可放 ...
- Fire Net(HDU-1045)(匈牙利最大匹配)(建图方式)
题意 有一个 n*n 的图,. 代表空白区域,X 代表墙,现在要在空白区域放置结点,要求同一行同一列只能放一个,除非有墙阻隔,问最多能放多少个点 思路 只有在墙的阻隔情况下,才会出现一行/列出现多个点 ...
随机推荐
- FSG压缩壳和ImportREC的使用 - 脱壳篇05
FSG压缩壳和ImportREC的使用 - 脱壳篇05 让编程改变世界 Change the world by program FSG这个壳可以说是有点儿不守妇道,尼玛你说你一个压缩壳就实现压缩功能得 ...
- debian gnome 3插件
1.gnome 配置-安装插件 http://maxubuntu.blogspot.com/2012/09/debian-gnome3.html hunagqf|hunaqf2|hunaqf3 2.快 ...
- UltraEdit的语法高亮显示配置
今天吴同学看到我电脑中有UltraEdit好奇地问我会不会用,我那个汗啊,不会用我装它干什么啊?其实当时装UltraEdit主要是用来写Java的,没有想到,工作一忙顾及不上学习Java的事情了.于是 ...
- SQL Server 索引的图形界面操作 <第十二篇>
一.索引的图形界面操作 SQL Server非常强大的就是图形界面操作.关于索引方面也一样那么强大,很多操作比如说重建索引啊,查看各种统计信息啊,都能够通过图形界面快速查看和操作,下面来看看SQL S ...
- cf471A MUH and Sticks
A. MUH and Sticks time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Struts2使用Interceptor实现权限控制的应用实例详解
Struts2使用Interceptor实现权限控制的应用实例详解 拦截器:是Struts2框架的核心,重点之重.因此,对于我们要向彻底学好Struts2.0.读源码和使用拦截器是必不可少的.少说了. ...
- Js apply 方法 具体解释
Js apply方法具体解释 我在一開始看到javascript的函数apply和call时,很的模糊,看也看不懂,近期在网上看到一些文章对apply方法和call的一些演示样例,总算是看的有点眉目了 ...
- 深入浅出:重温JAVA中接口与抽象的区别
抽象类:声明一个抽象类,就是在类的声明开头.在Class关键字的前面使用关键字abstract 下面定义一个抽象类,代码如下: abstract class A{ abstract void call ...
- 看漫画,学 Redux
Flux 架构已然让人觉得有些迷惑,而比 Flux 更让人摸不着头脑的是 Flux 与 Redux 的区别.Redux 是一个基于 Flux 思想的新架构方式,本文将探讨它们的区别. 如果你还没有看过 ...
- unity tips
1.在unity 的mecanim中,如果一个动画指向两个或两个以上的动画,那么在inspector中,transitions中可以看到所有的过渡路径,这些路径是有先后顺序的.