【题目】

Description

Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall should not come closer to the castle than a certain distance. If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will loose his head. Moreover, he demanded Architect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall. 

Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King's requirements.

The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet.

Input

The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King's castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle.

Next N lines describe coordinates of castle's vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.

Output

Write to the output file the single number that represents the minimal possible length of the wall in feet that could be built around the castle to satisfy King's requirements. You must present the integer number of feet to the King, because the floating numbers are not invented yet. However, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates.

Sample Input

9 100
200 400
300 400
300 300
400 300
400 400
500 400
500 200
350 200
200 200

Sample Output

1628
代码如下:
 #include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define Maxn 1010 struct Point
{
double x,y;
Point(double x=,double y=):x(x),y(y) {}
}; typedef Point Vector;
Point p[Maxn]; int n,l;
const double pi=3.1415926535898; double myabs(double x) {return x<?-x:x;} const double eps=1e-;
int dcmp(double x)//判断正负和零
{
if(myabs(x)<eps) return ;
else return x<?-:;
}
Vector operator - (Point A,Point B) {return Vector(A.x-B.x,A.y-B.y);} double Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}
double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}
double Length(Vector A) {return sqrt(Dot(A,A));} bool cmp(Point x,Point y) {return dcmp(x.x-y.x)==?(x.y<y.y):(x.x<y.x);}
Point ch[Maxn];int len=; void covexhull()
{
for(int i=;i<=n;i++)
{
while(len>&&Cross(ch[len]-ch[len-],p[i]-ch[len-])<=) len--;
ch[++len]=p[i];
}int k=len;
for(int i=n-;i>=;i--)
{
while(len>k&&Cross(ch[len]-ch[len-],p[i]-ch[len-])<=) len--;
ch[++len]=p[i];
}
if(n>) len--;
} double PolygonLength()
{
double L=;
for(int i=;i<=len;i++)
L+=Length(ch[i]-ch[i-]);
L+=Length(ch[]-ch[len]);
return L;
} int main()
{
scanf("%d%d",&n,&l);
for(int i=;i<=n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
sort(p+,p++n,cmp);
covexhull();
printf("%.0f\n",PolygonLength()+pi*l*);
return ;
}

[POJ1113]


2016-04-28 19:32:02


												

【POJ1113】Wall(凸包)的更多相关文章

  1. POJ1113 Wall —— 凸包

    题目链接:https://vjudge.net/problem/POJ-1113 Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submis ...

  2. POJ1113:Wall (凸包算法学习)

    题意: 给你一个由n个点构成的多边形城堡(看成二维),按顺序给你n个点,相邻两个点相连. 让你围着这个多边形城堡建一个围墙,城堡任意一点到围墙的距离要求大于等于L,让你求这个围墙的最小周长(看成二维平 ...

  3. POJ1113:Wall (凸包:求最小的多边形,到所有点的距离大于大于L)

    Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the ...

  4. POJ1113 Wall 凸包

    题目大意:建立围墙将城堡围起来,要求围墙至少距离城堡L,拐角处用圆弧取代,求围墙的长度. 题目思路:围墙长度=凸包周长+(2*PI*L),另外不知道为什么C++poj会RE,G++就没问题. #inc ...

  5. POJ1113 Wall

    题目来源:http://poj.org/problem?id=1113题目大意: 如图所示,给定N个顶点构成的一个多边形和一个距离值L.建立一个围墙,把这个多边形完全包含在内,且围墙距离多边形任一点的 ...

  6. [poj1113][Wall] (水平序+graham算法 求凸包)

    Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall ...

  7. POJ-1113 Wall 计算几何 求凸包

    题目链接:https://cn.vjudge.net/problem/POJ-1113 题意 给一些点,求一个能够包围所有点且每个点到边界的距离不下于L的周长最小图形的周长 思路 求得凸包的周长,再加 ...

  8. POJ1113 Wall【凸包】

    题意: 求把城堡围起来需要的最小墙壁周长. 思路: 围墙周长为=n条平行于凸包的线段+n条圆弧的长度=凸包周长+围墙离城堡距离L为半径的圆周长. 代码: ...还是看大佬写的,自己做个记录方便日后复习 ...

  9. POJ 1113 - Wall 凸包

    此题为凸包问题模板题,题目中所给点均为整点,考虑到数据范围问题求norm()时先转换成double了,把norm()那句改成<vector>压栈即可求得凸包. 初次提交被坑得很惨,在GDB ...

  10. hdu 1348 Wall (凸包)

    Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submis ...

随机推荐

  1. QT实现多语言切换

    功能需求: 网盘客户端要能够实现多国语言的切换,第一版要支持中.英文的切换.在实现过程中感觉QT对多国语言的支持还是很不错的,制作多语言包很方便,切换的逻辑也很简单.下面就来看一下QT中如何制作多语言 ...

  2. FileZilla命令行实现文件上传以及CreateProcess实现静默调用

    应用需求:         用户在选择渲染作业时面临两种情况:一是选择用户远程存储上的文件:二是选择本地文件系统中的文件进行渲染.由于渲染任务是在远程主机上进行的,实际进行渲染时源文件也是在ftp目录 ...

  3. 一个类搞定UIScrollView那些事

    前言 UIScrollView可以说是我们在日常编程中使用频率最多.扩展性最好的一个类,根据不同的需求和设计,我们都能玩出花来,当然有一些需求是大部分应用通用的,今天就聊一下以下需求,在一个categ ...

  4. Android自定义DataTimePicker(日期选择器)

    实现的效果就是在同一个布局上显示日期选择和时间选择,时间不准确bug修复 1.自定义类DateTimePickDialogUtil.java public class DateTimePickDial ...

  5. linux mysql 授权以及 iptables开启3306

    mysql授权ip段访问mysql grant all privileges on *.* to 'yang'@'192.168.1.%' identified by '123456'; linux ...

  6. web前端:html

    一.理解表单的作用 1.web 应用程序不仅仅是给用户显示数据,还应该给用户提供一个可以输入数据的图形用户界面.表单的主要作用在于在网页上提供一个图形用户界面,已采集和提交用户输入的数据. 2.htm ...

  7. mysql免安装版使用

    打开命令行,到bin目录下,输入net start mysql 启动服务,输入mysql -u root -p回车后输入密码,进入mysql.

  8. jQuery 取值、赋值的基本方法【转藏】

    /*获得TEXT.AREATEXT的值*/ var textval = $("#text_id").attr("value"); //或者 var textva ...

  9. 覆盖(override)和重载(overload)

    覆盖(override)重写和 重载(overload) 继承,重写--->多态   我懂了,你懂吗 ,不看看文章 java 子类重写父类的方法应注意的问题 Java多态性理解

  10. UI层调用WCF服务实例(源码)

    WCF原理性的东西,暂时还没有深入研究,只是在公司的项目中使用到了,会调用,然后再多做了一些了解,现在将它抽出来了一个小实例,写了一个WCF的demo. 我写的这个WCF.Demo主要包括数据契约和服 ...