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,判断平移后的直线的半平面交是否为空.

模板是照着敲的,还有一些地方不是很懂, 应该还要慢慢体会吧

&代码:

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std; struct Point {
double x, y;
Point(double x = 0, double y = 0): x(x), y(y) {}
}; typedef Point Vector; Vector operator + (const Vector& A, const Vector& B) {return Vector(A.x + B.x, A.y + B.y);}
Vector operator - (const Vector& A, const Vector& B) {return Vector(A.x - B.x, A.y - B.y);}
Vector operator * (const Vector& A, double p) {return Vector(A.x * p, A.y * p);}
double Dot(const Vector& A, const Vector& B) {return A.x * B.x + A.y * B.y;}
double Cross(const Vector& A, const Vector& B) {return A.x * B.y - A.y * B.x;}
double Length(const Vector& A) {return sqrt(Dot(A, A));}
Vector Normal(const Vector& A) {double l = Length(A); return Vector(-A.y / l, A.x / l);} double PolygonArea(vector<Point> p) {
int n = p.size();
double s = 0;
for(int i = 1; i < n - 1; i++) {
s += Cross(p[i] - p[0], p[i + 1] - p[0]);
}
return s / 2;
} struct Line {
Point p, v;
double ang;
Line() {}
Line(Point p, Vector v): p(p), v(v) {ang = atan2(v.y, v.x);}
bool operator < (const Line& l) const {
return ang < l.ang;
}
}; bool OnLeft(const Line& L, const Point& p) {
return Cross(L.v, p - L.p) > 0;
} Point GetLineIntersection(const Line& a, const 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;
} const 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), ans;
vector<Line> que(n);
que[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++;
que[++last] = L[i];
if(fabs(Cross(que[last].v, que[last - 1].v)) < eps) {
last--;
if(OnLeft(que[last], L[i].p)) que[last] = L[i];
}
if(first < last) p[last - 1] = GetLineIntersection(que[last - 1], que[last]);
}
while(first < last && !OnLeft(que[first], p[last - 1])) last--;
if(last - first <= 1) return ans;
p[last] = GetLineIntersection(que[last], que[first]);
for(int i = first; i <= last; i++)
ans.push_back(p[i]);
return ans;
} int main() {
//("E:1.in", "r", stdin);
int n;
while(scanf("%d", &n) == 1 && n) {
vector<Vector> p, v, nor;
int m, x, y;
for(int i = 0; i < n; i++) {
scanf("%d%d", &x, &y);
p.push_back(Point(x, y));
}
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]);
nor.push_back(Normal(v[i]));
}
double left = 0, right = 20000;
while(right - left > 1e-6) {
vector<Line> L;
double mid = left + (right - left) / 2;
for(int i = 0; i < n; i++)
L.push_back(Line(p[i] + nor[i]*mid, v[i]));
vector<Point> poly = HalfplaneIntersection(L);
if(poly.empty()) right = mid;
else left = mid;
}
printf("%f\n", left);
}
return 0;
}

LA 3890 Most Distant Point from the Sea(半平面交)的更多相关文章

  1. 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   ...

  2. 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 ...

  3. 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. /******* ...

  4. POJ3525 Most Distant Point from the Sea(半平面交)

    给你一个凸多边形,问在里面距离凸边形最远的点. 方法就是二分这个距离,然后将对应的半平面沿着法向平移这个距离,然后判断是否交集为空,为空说明这个距离太大了,否则太小了,二分即可. #pragma wa ...

  5. uvalive 3890 Most Distant Point from the Sea

    题意:求一个凸多边形中一点到边的最大距离. 思路:转换成在多边形内部,到每边距离为d的直线所围成的内多边形是否存在.也就是,二分距离+半平面交. #include<cstdio> #inc ...

  6. UVA 3890 Most Distant Point from the Sea(二分法+半平面交)

    题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=11358 [思路] 二分法+半平面交 二分与海边的的距离,由法向量可 ...

  7. UVALive 3890 Most Distant Point from the Sea(凸包最大内接园)

    一个n个点的凸多边形,求多边形中离多边形边界最远的距离.实际上就是求凸包最大内接圆的半径. 利用半平面交求解,每次二分枚举半径d,然后将凸包每条边所代表的半平面沿其垂直单位法向量平移d,看所有平移后的 ...

  8. 【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 ...

  9. 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   ...

随机推荐

  1. centos7配置vsftpd

    ftp服务器192.168.1.198 1.关闭selinux.关闭防火墙 2.yum -y install vsftpd* 3.开启服务,通过浏览器访问ftp://192.168.1.198,匿名用 ...

  2. 免费SSL证书Let's Encrypt(certbot)安装使用教程

    免费SSL证书Let's Encrypt(certbot)安装使用教程 https://www.vpser.net/build/letsencrypt-certbot.html

  3. HTML基础标签大全

    HTML 标签大全及属性 常用的块级元素有 :div , from , table, p ,pre,h1-h5,dl,ol,ul 常用的内联元素:span ,a ,strong,em ,label , ...

  4. ios开发dismiss所有控制器

    -(void)dismissToRootViewController { UIViewController *vc = self; while (vc.presentingViewController ...

  5. java常用工具所在的包

    org.apache.commons.lang3:1)StringUtils.isBlank org.springframework.util:1)ResourceUtils.getFile(&quo ...

  6. redis的基本介绍

    redis是什么? redis是一种菲关系型数据库,存储key-value类型的数据. redis支持的数据类型 这里所说的数据类型其实就是value对应的数据类型.一共有五种: String 1.S ...

  7. activeMQ的安装和使用

    什么是ActiveMQ? 一款开源的JMS具体实现,是一个易于使用的消息中间件,一个消息容器 安装 下载 官方网站:http://activemq.apache.org/ 解压 linux下的安装,解 ...

  8. java 网络编程(四)TCP通讯

    客户端: package cn.sasa.TcpDemo; import java.io.IOException; import java.io.InputStream; import java.io ...

  9. 洛谷P3237 米特运输 [HNOI2014] hash/二进制分解

    正解:hash/二进制分解 解题报告: 传送门! umm首先提取下题意趴QAQ 大概是说给一棵树,每个点有一个权值,要求修改一些点的权值,使得同一个父亲的儿子权值相同,且父亲的权值必须是所有儿子权值之 ...

  10. linux网络设置和虚拟机克隆转移之后网卡找不到

    linux网络设置和虚拟机克隆转移之后Error:No suitable device found:no device found for connection 'System eth0'问题解决   ...