zoj1002
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
直接上代码吧
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <list>
#define OPEN 1
#define CASTLE 2
#define BLOCK 3
#define WALL 4
// summary: 用贪心原理找到一个影响地图最小的城堡的位置
// param in: map[] 输入的地图. n 地图大小.
// param out: min_x,min_y 得出的放置的城堡的x,y坐标
// return: true 找到了合适的位置. false 没有可以放置城堡的地方了
#define MAX_POINT 10000 // 哨兵数
bool FindMinInfluencePosition(int map[][], int n, int &min_x, int &min_y)
{
int min_point = MAX_POINT; // 赋哨兵值
min_x = min_y = ; for (int y = ; y < n; y++)
{
for (int x = ; x < n; x++)
{
if ( map[y][x] == OPEN)
{
int curr_point = ; // 向上
for (int j = y - ; j >= && map[j][x] != WALL; j--)
{
if ( map[j][x] == OPEN)
curr_point++;
} // 向下
for (int j = y + ; j < n && map[j][x] != WALL; j++)
{
if ( map[j][x] == OPEN)
curr_point++;
} // 向左
for (int j = x - ; j >= && map[y][j] != WALL; j--)
{
if ( map[y][j] == OPEN)
curr_point++;
} // 向右
for (int j = x + ; j < n && map[y][j] != WALL; j++)
{
if ( map[y][j] == OPEN)
curr_point++;
} if (curr_point < min_point)
{
min_point = curr_point;
min_x = x;
min_y = y;
}
}
}
} // 检测是否放置了城堡
if (min_point != MAX_POINT)
return true;
else
return false; } // summary: 根据位置放置城堡,在地图上标记出来
// param in: map[] 输入的地图. n 地图大小. x, y 放置的城堡的位置
void PlaceCastle(int map[][], int n, int x, int y)
{
map[y][x] = CASTLE; // 向上
for (int j = y - ; j >= && map[j][x] != WALL; j--)
{
map[j][x] = BLOCK;
} // 向下
for (int j = y + ; j < n && map[j][x] != WALL; j++)
{
map[j][x] = BLOCK;
} // 向左
for (int j = x - ; j >= && map[y][j] != WALL; j--)
{
map[y][j] = BLOCK;
} // 向右
for (int j = x + ; j < n && map[y][j] != WALL; j++)
{
map[y][j] = BLOCK;
}
} int main ()
{
int map[][];
std::list<int> answer;//构造函数
int n;
char c; // Read the data
std::cin>>n;
while (n != )
{
// Read map data
for (int y = ; y < n; y++)
{
for (int x = ; x < n; x++)
{
std::cin>>c;
if (c == '.')
map[y][x] = OPEN;
else
map[y][x] = WALL;
}
} // Processing maps
int castle_number = ;
int min_x, min_y;
while (FindMinInfluencePosition(map, n, min_x, min_y) == true )
{
castle_number++;
PlaceCastle(map, n, min_x, min_y);
} // Record the data
answer.push_back(castle_number); // Enter the next number
std::cin>>n;
} // The number of output
for (std::list<int>::iterator it=answer.begin() ; it != answer.end(); ++it)
std::cout <<*it<<"\n"; return ;
} //菜鸟一枚,代码来自大神的博客,只是记录自己的学习,具体那一篇后来没找到,若有侵犯,楼主请留言
zoj1002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 ...
- zoj1002 Fire Net
Fire Net Time Limit: 2 Seconds Memory Limit: 65536 KB Suppose that we have a square city with s ...
- C++二分图匹配基础:zoj1002 FireNet 火力网
直接给出题目吧... 问题 D(1988): [高级算法]火力网 时间限制: 1 Sec 内存限制: 128 MB 题目描述 给出一个N*N的网格,用'.'表示空地,用'X'表示墙.在网格上放碉堡,可 ...
- OJ题目分类
POJ题目分类 | POJ题目分类 | HDU题目分类 | ZOJ题目分类 | SOJ题目分类 | HOJ题目分类 | FOJ题目分类 | 模拟题: POJ1006 POJ1008 POJ1013 P ...
- 牛人的ACM经验 (转)
一:知识点 数据结构: 1,单,双链表及循环链表 2,树的表示与存储,二叉树(概念,遍历)二叉树的 应用(二叉排序树,判定树,博弈 ...
- ACM算法锦集
一:知识点 数据结构: 1,单,双链表及循环链表 2,树的表示与存储,二叉树(概念,遍历)二叉树的 应用(二叉排序树,判定树,博弈树,解答树等) 3,文件操作(从文本文件中读入数据并输出到文本文 件中 ...
随机推荐
- electron-edge-js 环境搭建
确保nodejs环境 为 10.X (因为edge的编译需要node对应版本的支持,太新的node不包含对应edge的编译) 1.创建工程2.使用npm init初始化程序信息3.使用npm ins ...
- CSS3-背景(background-image、background-size、background-origin、background-clip)
CSS3中新的背景属性:background-image.background-size.background-origin.background-clip 背景图片:background-image ...
- js中for循环(原生js)
1,普通for循环,经常用的数组遍历 var arr = [1,2,3,4,5]; for ( var i = 0; i <arr.length; i++){ console.log(arr[i ...
- 提高unigui开发效率的两个方法(02)
1.编译时自己退出运行的程序. 在做unigui开发时,每次编译运行时,unigui的应用都会在后台运行,每次重新编译时都必须手工在任务栏里将应用退出才行,非常麻烦,可以在项目编译的参数里加上杀进程的 ...
- Numpy 为运算
Numpy “bitwise_” 开头的函数是位运算函数: Numpy 位运算包括以下几个函数: 函数 描述 bitwise_and 对数组元素执行位与操作 bitwise_or 对数组元素执行 ...
- <每日一课学习笔记> "mysql为什么加了索引还是慢查询"
long_query_time mysql判断慢查询的依据是,sql执行时间与系统参数 long_query_time 作比较,如果大于这个参数,就会将这条sql计入慢查询语句中.long_query ...
- VirtualBox安装Debian
1.下载Debian的dvd1,按照http://www.jb51.net/os/85858.html网上教程安装Debian 1.1.我创建了20G的虚拟磁盘,分区的时候我分了3个区,2G交换空间, ...
- boost::program_options 解析命令行参数
源码: #include <boost/program_options.hpp> namespace po = boost::program_options; int main(int a ...
- boost::timer demo
#include <iostream> #include <boost/timer.hpp> //timer的头文件 using namespace boost; //打开bo ...
- ch5 创建类似按钮的链接
锚是行内元素,所以只有在单击链接的内容时它们才会激活,如果可以实现为按钮的效果,就可以有更大的可单击区域,实现方法为:display设置为block,修改width.height和其他属性,代码如下: ...