poj 3525 求凸包的最大内切圆
Time Limit: 5000MS | Memory Limit: 65536K | |||
Total Submissions: 3640 | Accepted: 1683 | Special Judge |
Description
The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.
In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.
Input
The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.
n | ||
x1 | y1 | |
⋮ | ||
xn | yn |
Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.
n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi)–(xi+1, yi+1) (1 ≤ i ≤ n − 1) and the line segment (xn,yn)–(x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.
You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.
The last dataset is followed by a line containing a single zero.
Output
For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10−5). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
Sample Input
4
0 0
10000 0
10000 10000
0 10000
3
0 0
10000 0
7000 1000
6
0 40
100 20
250 40
250 70
100 90
0 70
3
0 0
10000 10000
5000 5001
0
Sample Output
5000.000000
494.233641
34.542948
0.353553 题目大意:求凸包的最大内切圆
分析:二分找答案,对凸包的每条边平移收缩,若半平面交不为空集,则可以存在此半径的圆。
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
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 Point& A, const Point& 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 area = 0;
for(int i = 1; i < n-1; i++)
area += Cross(p[i]-p[0], p[i+1]-p[0]);
return area/2;
} // 有向直线。它的左边就是对应的半平面
struct Line {
Point P; // 直线上任意一点
Vector v; // 方向向量
double ang; // 极角,即从x正半轴旋转到向量v所需要的角(弧度)
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;
}
}; // 点p在有向直线L的左边(线上不算)
bool OnLeft(const Line& L, const Point& p) {
return Cross(L.v, p-L.P) > 0;
} // 二直线交点,假定交点惟一存在
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; // 半平面交主过程
vector<Point> HalfplaneIntersection(vector<Line> L) {
int n = L.size();
sort(L.begin(), L.end()); // 按极角排序
int first, last,i; // 双端队列的第一个元素和最后一个元素的下标
vector<Point> p(n); // p[i]为q[i]和q[i+1]的交点
vector<Line> q(n); // 双端队列
vector<Point> ans; // 结果
q[first=last=0] = L[0]; // 双端队列初始化为只有一个半平面L[0]
for(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) { // 两向量平行且同向,取内侧的一个
last--;
if(OnLeft(q[last], L[i].P)) q[last] = L[i];
}
if(first < last) p[last-1] = GetLineIntersection(q[last-1], q[last]);
}
while(first < last && !OnLeft(q[first], p[last-1])) last--; // 删除无用平面
if(last - first <= 1) return ans; // 空集
p[last] = GetLineIntersection(q[last], q[first]); // 计算首尾两个半平面的交点
// 从deque复制到输出中
for(i = first; i <= last; i++) ans.push_back(p[i]);
return ans;
} int main() {
int n;
while(scanf("%d", &n) == 1 && n)
{
vector<Vector> p, v, normal;
int i, x, y;
for(i = 0; i < n; i++) { scanf("%d%d", &x, &y); p.push_back(Point(x,y)); }
for(i = 0; i < n; i++) {
v.push_back(p[(i+1)%n]-p[i]);
normal.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(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;
}
poj 3525 求凸包的最大内切圆的更多相关文章
- POJ 2187 求凸包上最长距离
简单的旋转卡壳题目 以每一条边作为基础,找到那个最远的对踵点,计算所有对踵点的点对距离 这里求的是距离的平方,所有过程都是int即可 #include <iostream> #includ ...
- 计算几何--求凸包模板--Graham算法--poj 1113
Wall Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 28157 Accepted: 9401 Description ...
- poj 1113:Wall(计算几何,求凸包周长)
Wall Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 28462 Accepted: 9498 Description ...
- POJ 3348 Cows 凸包 求面积
LINK 题意:给出点集,求凸包的面积 思路:主要是求面积的考察,固定一个点顺序枚举两个点叉积求三角形面积和除2即可 /** @Date : 2017-07-19 16:07:11 * @FileNa ...
- POJ 2187 Beauty Contest【旋转卡壳求凸包直径】
链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#probl ...
- 简单几何(求凸包点数) POJ 1228 Grandpa's Estate
题目传送门 题意:判断一些点的凸包能否唯一确定 分析:如果凸包边上没有其他点,那么边想象成橡皮筋,可以往外拖动,这不是唯一确定的.还有求凸包的点数<=2的情况一定不能确定. /********* ...
- POJ 1113 Wall(Graham求凸包周长)
题目链接 题意 : 求凸包周长+一个完整的圆周长. 因为走一圈,经过拐点时,所形成的扇形的内角和是360度,故一个完整的圆. 思路 : 求出凸包来,然后加上圆的周长 #include <stdi ...
- POJ 1113 Wall 凸包求周长
Wall Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 26286 Accepted: 8760 Description ...
- Wall - POJ 1113(求凸包)
题目大意:给N个点,然后要修建一个围墙把所有的点都包裹起来,但是要求围墙距离所有的点的最小距离是L,求出来围墙的长度. 分析:如果没有最小距离这个条件那么很容易看出来是一个凸包,然后在加上一个最小距离 ...
随机推荐
- 使用JOSM编辑OpenStreetMap地图
申明:转载请注明出处! 网上关于JOSM的使用大多只介绍了如何安装和优缺点,对于我这种小白完全还是不会,于是Google了一番,国外关于JOSM的使用的文章还是很多的, 选中一篇讲解的非常详细来翻译, ...
- [Batch檔案筆記] 在UNC路徑中執行Batch檔
為了讓其他人可以免安裝又可以執行python程式所以我把python portable版本 winpython 放在samba的空間共享但是使用者如果要開 winpython cammand prom ...
- Active Directory网域
Active Directory网域 3.1Windows网络的管理方式 3.1.1工作组模式 工作组由一组用网络连接在一起的计算机组成,他们将计算机内的资源共享给用户访问.工作组网络也被称为“对等式 ...
- Bootstrap 网格系统(Grid System)实例2
Bootstrap 网格系统(Grid System):堆叠水平,两种样式 <!DOCTYPE html><html><head><meta http-equ ...
- Spring框架针对dao层的jdbcTemplate操作crud之query查询数据操作
查询目标是完成3个功能: (1)查询表,返回某一个值.例如查询表中记录的条数,返回一个int类型数据 (2)查询表,返回结果为某一个对象. (3)查询表,返回结果为某一个泛型的list集合. 一.查询 ...
- ios之UIToolBar
toolbar除了可以和navigationController一起用之外,也可以独立用到view里.工具栏UIToolbar – 一般显示在底部,用于提供一组选项,让用户执行一些功能,而并非用于在完 ...
- POJ 3080 Blue Jeans、POJ 3461 Oulipo——KMP应用
题目:POJ3080 http://poj.org/problem?id=3080 题意:对于输入的文本串,输出最长的公共子串,如果长度相同,输出字典序最小的. 这题数据量很小,用暴力也是16ms,用 ...
- 学习笔记(_huaji_)
假如我没有见过太阳,我也许会忍受黑暗. 如果我知道自己会在哪里死去,我就永远都不去那儿.失败的经历,其实也有它的价值. 人的过失会带来错误,但要制造真正的灾难还得用计算机. 嘴角微微上扬已不复当年轻狂 ...
- Bluefruit LE Sniffer - Bluetooth Low Energy (BLE 4.0) - nRF51822 驱动安装及使用
BLE Sniffer https://www.adafruit.com/product/2269 Bluefruit LE Sniffer - Bluetooth Low Energy (BLE 4 ...
- jQuery和Vue
jQuery 概述 是js的一种函数库有美国人 John Resig编写 特点 写的少,做的多,国内用的jq1.0版本,可以兼容低版本的浏览器,支持链式编程或链式调用和隐式迭代 链式编程 $(this ...