Fire Net(深度优先搜索)
Time Limit: 2 Seconds Memory Limit: 65536 KB
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 legalprovided 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.
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.
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
Source: Zhejiang University Local Contest 2001
#include <iostream>
#include <cstdio> using namespace std; const int SZ = ;
int n;
char map[SZ][SZ];
bool vis[SZ][SZ]; int DFS(int y, int x)
{
if(y == n) return ;
if(map[y][x] == 'X')
{
if(x < n - ) return DFS(y, x + );
else return DFS(y + , );
}
bool ok = true;
for(int i = x - ; i > -; i--)
{
if(map[y][i] == 'X') break;
else if(vis[y][i])
{
ok = false;
break;
}
}
if(ok)
{
for(int i = y - ; i > -; i--)
{
if(map[i][x] == 'X') break;
else if(vis[i][x])
{
ok = false;
break;
}
}
}
if(ok)
{
int add0 = , add1 = ;
if(x == n - )
add0 += DFS(y + , );
else
add0 += DFS(y, x + );
vis[y][x] = true;
if(x == n - )
add1 += DFS(y + , );
else
add1 += DFS(y, x + );
vis[y][x] = false;
return add0 > add1 ? add0 : add1;
}
else
{
if(x < n - ) return DFS(y, x + );
else return DFS(y + , );
}
} int main()
{
char ch;
while(scanf("%d", &n) && n)
{
for(int i = ; i < n; i++)
{
scanf("%s", map[i]);
}
memset(vis, false, sizeof(vis));
printf("%d\n", DFS(, ));
}
return ;
}
Fire Net(深度优先搜索)的更多相关文章
- ZOJ1002 —— 深度优先搜索
ZOJ1002 —— Fire net Time Limit: 2000 ms Memory Limit: 65536 KB Suppose that we have a square city wi ...
- 深度优先搜索(DFS)
[算法入门] 郭志伟@SYSU:raphealguo(at)qq.com 2012/05/12 1.前言 深度优先搜索(缩写DFS)有点类似广度优先搜索,也是对一个连通图进行遍历的算法.它的思想是从一 ...
- 初涉深度优先搜索--Java学习笔记(二)
版权声明: 本文由Faye_Zuo发布于http://www.cnblogs.com/zuofeiyi/, 本文可以被全部的转载或者部分使用,但请注明出处. 上周学习了数组和链表,有点基础了解以后,这 ...
- 挑战程序2.1.4 穷竭搜索>>深度优先搜索
深度优先搜索DFS,从最开始状态出发,遍历一种状态到底,再回溯搜索第二种. 题目:POJ2386 思路:(⊙v⊙)嗯 和例题同理啊,从@开始,搜索到所有可以走到的地方,把那里改为一个值(@或者 ...
- 回溯 DFS 深度优先搜索[待更新]
首先申明,本文根据微博博友 @JC向北 微博日志 整理得到,本文在这转载已经受作者授权! 1.概念 回溯算法 就是 如果这个节点不满足条件 (比如说已经被访问过了),就回到上一个节点尝试别 ...
- 总结A*,Dijkstra,广度优先搜索,深度优先搜索的复杂度比较
广度优先搜索(BFS) 1.将头结点放入队列Q中 2.while Q!=空 u出队 遍历u的邻接表中的每个节点v 将v插入队列中 当使用无向图的邻接表时,复杂度为O(V^2) 当使用有向图的邻接表时, ...
- 深度优先搜索(DFS)
定义: (维基百科:https://en.wikipedia.org/wiki/Depth-first_search) 深度优先搜索算法(Depth-First-Search),是搜索算法的一种.是沿 ...
- 图的遍历之深度优先搜索(DFS)
深度优先搜索(depth-first search)是对先序遍历(preorder traversal)的推广.”深度优先搜索“,顾名思义就是尽可能深的搜索一个图.想象你是身处一个迷宫的入口,迷宫中的 ...
- HDU 1241 Oil Deposits DFS(深度优先搜索) 和 BFS(广度优先搜索)
Oil Deposits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...
- 算法总结—深度优先搜索DFS
深度优先搜索(DFS) 往往利用递归函数实现(隐式地使用栈). 深度优先从最开始的状态出发,遍历所有可以到达的状态.由此可以对所有的状态进行操作,或列举出所有的状态. 1.poj2386 Lake C ...
随机推荐
- RIGHT-BICEP单元测试——“二柱子四则运算升级版”
RIGHT-BICEP单元测试 ——“二柱子四则运算升级版” ”单元测试“这对于我们来说是一个全新的专业含义,在上了软件工程这门课,并当堂编写了简单的"求一组数中的最大值"函数的单 ...
- Java 数组转字符
public static String toString(int[] arr){ String temp = ""; for(int i = 0;i<arr.length; ...
- C++ Primer Plus学习:第十五章
第十五章 友元.异常和其他 友元 友元类 表 0-1 class Tv { public: friend class Remote; } Remote类可以使用Tv的数据成员,Remote类在Tv类后 ...
- "firstday"-软件工程
阅读以下文章 http://www.thea.cn/news/terminal/9/9389.html http://www.shzhidao.cn/system/2015/09/22/0102 ...
- beta阶段——项目复审
beta阶段--项目复审 小组的名字和链接 优点 缺点 bug 排名顺序 颜罗王team http://www.cnblogs.com/LDLYMteam 界面清新,音乐能够选择是否播放,词汇按照四六 ...
- linux mysql表名大小写
1.用ROOT登录,修改/etc/my.cnf 2.在[mysqld]下加入一行:lower_case_table_names=1 0:区分大小写,1:不区分大小写 3.重新启动数据库即可
- 第107天:Ajax 实现简单的登录效果
使用 Ajax 实现简单的登录效果 Ajax是一项使局部网页请求服务器信息,而不需整体刷新网页内容的异步更新技术.这使得向服务器请求的数据量大大减少,而且不会因局部的请求失败而影响到整体网页的加载. ...
- JSON字符串转换成对象时候 需要有默认构造器 因为这是通过反射创建的 反射是先通过默认构造器创建对象的
JSON字符串转换成对象时候 需要有默认构造器 因为这是通过反射创建的 反射是先通过默认构造器创建对象的
- cdq分治学习
看了stdcall大佬的博客 传送门: http://www.cnblogs.com/mlystdcall/p/6219421.html 感觉cdq分治似乎很多时候都要用到归并的思想
- 树状数组模板(pascal) 洛谷P3374 【模板】树状数组1
题目描述 如题,已知一个数列,你需要进行下面两种操作: 1.将某一个数加上x 2.求出某区间每一个数的和 输入输出格式 输入格式: 第一行包含两个整数N.M,分别表示该数列数字的个数和操作的总个数. ...