LA3890 Most Distant Point from the Sea
题意
分析
可以二分答案,检验就用半平面交,如果平面非空则合法。
时间复杂度\(O(T n \log^2 n)\)
代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
rg T data=0;
rg int w=1;
rg char ch=getchar();
while(!isdigit(ch))
{
if(ch=='-')
w=-1;
ch=getchar();
}
while(isdigit(ch))
{
data=data*10+ch-'0';
ch=getchar();
}
return data*w;
}
template<class T>T read(T&x)
{
return x=read<T>();
}
using namespace std;
typedef long long ll;
struct Point
{
double x,y;
Point(double x=0,double y=0)
:x(x),y(y){}
};
typedef Point Vector;
Vector operator+(co Vector&A,co Vector&B)
{
return Vector(A.x+B.x,A.y+B.y);
}
Vector operator-(co Vector&A,co Vector&B)
{
return Vector(A.x-B.x,A.y-B.y);
}
Vector operator*(co Vector&A,double p)
{
return Vector(A.x*p,A.y*p);
}
Vector operator/(co Vector&A,double p)
{
return Vector(A.x/p,A.y/p);
}
double Dot(co Vector&A,co Vector&B)
{
return A.x*B.x+A.y*B.y;
}
double Cross(co Vector&A,co Vector&B)
{
return A.x*B.y-A.y*B.x;
}
double Length(co Vector&A)
{
return sqrt(Dot(A,A));
}
Vector Normal(co Vector&A)
{
double L=Length(A);
return Vector(-A.y/L,A.x/L);
}
double PolygonArea(vector<Point> p)
{
int n=p.size();
double area=0;
for(int i=1;i<n-1;++i)
area+=Cross(p[i]-p[0],p[i+1]-p[0]);
return area/2;
}
struct Line
{
Point P;
Vector v;
Line(Point P=0,Vector v=0)
:P(P),v(v){}
double angle()co
{
return atan2(v.y,v.x);
}
bool operator<(co Line&rhs)co
{
return angle()<rhs.angle();
}
};
bool OnLeft(co Line&L,co Point&p)
{
return Cross(L.v,p-L.P)>0;
}
Point LineLineIntersection(co Line&a,co Line&b)
{
Vector u=a.P-b.P;
double t=Cross(b.v,u)/Cross(a.v,b.v);
return a.P+a.v*t;
}
co double INF=1e8;
co double eps=1e-6;
vector<Point>HalfplaneIntersection(vector<Line>L)
{
int n=L.size();
sort(L.begin(),L.end());
int first,last;
vector<Point>p(n);
vector<Line>q(n);
vector<Point>ans;
q[first=last=0]=L[0];
for(int i=1;i<n;++i)
{
while(first<last&&!OnLeft(L[i],p[last-1]))
--last;
while(first<last&&!OnLeft(L[i],p[first]))
++first;
q[++last]=L[i];
if(fabs(Cross(q[last].v,q[last-1].v))<eps)
{
--last;
if(OnLeft(q[last],L[i].P))
q[last]=L[i];
}
if(first<last)
p[last-1]=LineLineIntersection(q[last-1],q[last]);
}
while(first<last&&!OnLeft(q[first],p[last-1]))
--last;
if(last-first<=1)
return ans;
p[last]=LineLineIntersection(q[last],q[first]);
for(int i=first;i<=last;++i)
ans.push_back(p[i]);
return ans;
}
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n;
while(read(n))
{
vector<Vector>p,v,normal;
for(int i=0;i<n;++i)
p.push_back(Point(read<int>(),read<int>()));
if(PolygonArea(p)<0)
reverse(p.begin(),p.end());
for(int i=0;i<n;++i)
{
v.push_back(p[(i+1)%n]-p[i]);
normal.push_back(Normal(v[i]));
}
double left=0,right=2e4;
while(right-left>eps)
{
vector<Line>L;
double mid=(left+right)/2;
for(int i=0;i<n;++i)
L.push_back(Line(p[i]+normal[i]*mid,v[i]));
vector<Point>poly=HalfplaneIntersection(L);
if(poly.empty())
right=mid;
else
left=mid;
}
printf("%.6lf\n",left);
}
return 0;
}
LA3890 Most Distant Point from the Sea的更多相关文章
- POJ 3525 Most Distant Point from the Sea [半平面交 二分]
Most Distant Point from the Sea Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 5153 ...
- LA 3890 Most Distant Point from the Sea(半平面交)
Most Distant Point from the Sea [题目链接]Most Distant Point from the Sea [题目类型]半平面交 &题解: 蓝书279 二分答案 ...
- 【POJ】【3525】Most Distant Point from the Sea
二分+计算几何/半平面交 半平面交的学习戳这里:http://blog.csdn.net/accry/article/details/6070621 然而这题是要二分长度r……用每条直线的距离为r的平 ...
- POJ 3525 Most Distant Point from the Sea (半平面交+二分)
Most Distant Point from the Sea Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 3476 ...
- POJ 3525/UVA 1396 Most Distant Point from the Sea(二分+半平面交)
Description The main land of Japan called Honshu is an island surrounded by the sea. In such an isla ...
- POJ3525-Most Distant Point from the Sea(二分+半平面交)
Most Distant Point from the Sea Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 3955 ...
- POJ 3525 Most Distant Point from the Sea (半平面交)
Description The main land of Japan called Honshu is an island surrounded by the sea. In such an isla ...
- 【POJ 3525】Most Distant Point from the Sea(直线平移、半平面交)
按逆时针顺序给出n个点,求它们组成的多边形的最大内切圆半径. 二分这个半径,将所有直线向多边形中心平移r距离,如果半平面交不存在那么r大了,否则r小了. 平移直线就是对于向量ab,因为是逆时针的,向中 ...
- 简单几何(半平面交+二分) LA 3890 Most Distant Point from the Sea
题目传送门 题意:凸多边形的小岛在海里,问岛上的点到海最远的距离. 分析:训练指南P279,二分答案,然后整个多边形往内部收缩,如果半平面交非空,那么这些点构成半平面,存在满足的点. /******* ...
随机推荐
- JavaWeb请求中文乱码
解决中文乱麻问题,页面端发出的数据作两次encodeURI var name="张三"; encodeURI(encodeURI(name)); 后台解码: URLDecoder. ...
- Python3.x:实现多任务(多进程)
Python3.x:实现多任务(多进程) # python3 # author lizm # datetime 2018-02-13 16:00:00 # -*- coding: utf-8 -*- ...
- kali安装后配置
0x00.安装Vmware Tools 由于是在VMware Workstation里面安装的,所以需要首先安装VMware tools工具方便我们Ctrl+C和Ctrl+V,步骤如下: 在VMWar ...
- ubuntu16.04的anacoda内置的spyder不支持中文【学习笔记】
执行下面的语句:将libfcitxplatforminputcontextplugin.so复制到anaconda2的安装目录下的platforminputcontexts目录重启生效 cp /usr ...
- Curator学习
Curator是对zookeeper的高级封装. 考虑到使用它来开发zookeeper应用的方便,特此来记录总结学习与开发使用过程的问题. 1. curator-framework Curator F ...
- centos 6的LAMP一键安装包(可选择/升级版本)
安装步骤 事前准备(安装 wget.screen.unzip,创建 screen 会话) yum -y install wget screen git git clone 并赋予脚本执行权限 git ...
- 电脑 HOST 文件
路径: C:\Windows\System32\drivers\etc\HOSTS
- centos 使用rz sz指令
在linux下安装rz很方便,使用 yum install lrzsz 就可以安装,正常使用rz和sz命令. 下面对sz和rz命令的一点介绍: 一般来说,linux服务器大多是通过ssh客户端来进行远 ...
- CodeChef CHEFSOC2 Chef and Big Soccer 水dp
Chef and Big Soccer Problem code: CHEFSOC2 Tweet ALL SUBMISSIONS All submissions for this prob ...
- ACM International Collegiate Programming Contest, Tishreen Collegiate Programming Contest (2018) Syria, Lattakia, Tishreen University, April, 30, 2018
ACM International Collegiate Programming Contest, Tishreen Collegiate Programming Contest (2018) Syr ...