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 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.
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
题目中译:
有n×n个格子,每个格子是堡垒或城墙或空地。
你原来知道哪里有城墙或空地。
堡垒会向东西南北四方向发射炮弹,城墙能阻挡炮弹。
问最多能放几个堡垒。
题目思维:
1.我们遍历每个格子,能放就放,并且将放完后与它有冲突的格子进行标记,最后输出。
可惜如果这样做不到最大优化。
2.我们仍然深度优先搜索每个格子,这次如果没有打过标记,则将格子标记0或1,如果标记过则只标记0,继续向下遍历,最后填满后更新最大值。
注意没打过表记的格子,打完标记要还原。
Code:
// by kyrence
#include <bits/stdc++.h>
using namespace std; int n, ans;
bool can[][];
char c; bool valid(int x, int y) {
return x > && y > && x <= n && y <= n;
} void FW(int x, int y, int dx, int dy) {
do {
can[x][y] = ;
x += dx;
y += dy;
} while (valid(x, y) && can[x][y]);
} void dfs(int dep, int tot) {
if (dep > n * n) {
ans = max(tot, ans);
return;
}
dfs(dep + , tot);
int x = (dep + n - ) / n;
int y = dep % n;
if (!y) y = n;
if (!can[x][y]) return;
bool cam[][];
for (int i = ; i <= n; i++)
for (int j = ; j <= n; j++)
cam[i][j] = can[i][j];
FW(x, y, , );
FW(x, y, -, );
FW(x, y, , );
FW(x, y, , -);
dfs(dep + , tot + );
for (int i = ; i <= n; i++)
for (int j = ; j <= n; j++)
can[i][j] = cam[i][j];
} int main() {
while (scanf("%d", &n) == && n) {
ans = -;
for (int i = ; i <= n; i++)
for (int j = ; j <= n; j++) {
cin >> c;
can[i][j] = (c == '.');
}
dfs(, );
printf("%d\n", ans);
}
return ;
}
std
ZOJ1002 —— 深度优先搜索的更多相关文章
- 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
- Python标准库之shelve模块(序列化与反序列化)
shelve模块是一个简单的key,value将内存数据通过文件持久化的模块,可以持久化任何picklel可支持的Python数据格式. 序列化 序列化源代码: import shelve impor ...
- 题解【AcWing274】移动服务
题面 非常好的优化 \(\text{DP}\) 状态表示的题目. 首先可以设 \(dp_{i,x,y,z}\) 表示已经做完了前 \(i\) 个请求,现在的 \(3\) 名服务员分别在 \(x\) . ...
- 题解【洛谷P1618】 三连击(升级版)
设三个数分别为n1.n2.n3,因为三个数的比为A:B:C,取一份量i,使得A·i=x,B·i=y,C·i=z(·是*的意思). 所以我们的代码只需要枚举i,并以此判断n1.n2.n3是否为三位数且包 ...
- Linux system 初步
快捷键: open a new terminal: ctrl+alt+T; close current terminal: ctrl+shift+W; switch windows: alt+tab ...
- Hibernate:Hibernate缓存策略详解
一:为什么使用Hibernate缓存: Hibernate是一个持久层框架,经常访问物理数据库. 为了降低应用程序访问物理数据库的频次,从而提高应用程序的性能. 缓存内的数据是对物理数据源的复制,应用 ...
- codeforces 1288C. Two Arrays(dp)
链接:https://codeforces.com/contest/1288/problem/C C. Two Arrays 题意:给定一个数n和一个数m,让构建两个数组a和b满足条件,1.数组中所有 ...
- 【STM32H7教程】第60章 STM32H7的DAC应用之定时器触发实现DMA方式双通道波形
完整教程下载地址:http://www.armbbs.cn/forum.php?mod=viewthread&tid=86980 第60章 STM32H7的DAC应用之定时器触发实 ...
- Java连载77-Integer常用方法、Integer、int、String三者相互转化、自动装箱、自动拆箱
一.关于Integer中常用的方法 package com.bjpowernode.java_learning; public class D77_1_ { public static void ...
- SpringBoot整合WEB开发--(三)文件上传
文件上传: Java中文件上传一共涉及到两个组件,CommonsMultipartResolver和StandardServletMultipartResolver,其中CommonsMultipar ...
- (ghrd)pio设置
设置中是下降沿,边沿触发.