uva 1396 - Most Distant Point from the Sea
半平面的交,二分的方法;
#include<cstdio>
#include<algorithm>
#include<cmath>
#define eps 1e-6
using namespace std; int dcmp(double x)
{
return fabs(x) < eps ? : (x > ? : -);
} struct Point
{
double x;
double y;
Point(double x = , double y = ):x(x), y(y) {}
};
typedef Point Vector; Vector operator + (Point A, Point B)
{
return Vector(A.x + B.x, A.y + B.y);
} Vector operator - (Point A, Point B)
{
return Vector(A.x - B.x, A.y - B.y);
} Vector operator * (Point A, double p)
{
return Vector(A.x * p, A.y * p);
} Vector operator / (Point A, double p)
{
return Vector(A.x / p, A.y / p);
}
double dot(Point a,Point b)
{
return a.x*b.x+a.y*b.y;
}
double cross(Point a,Point b)
{
return a.x*b.y-a.y*b.x;
} Vector nomal(Vector a)
{
double l=sqrt(dot(a,a));
return Vector(-a.y/l,a.x/l);
} struct line
{
Point p;
Vector v;
double ang;
line() {}
line(Point p,Vector v):p(p),v(v)
{
ang=atan2(v.y,v.x);
}
bool operator<(const line &t)const
{
return ang<t.ang;
}
}; bool onleft(line l,Point p)
{
return (cross(l.v,p-l.p)>);
} Point getintersection(line a,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;
} int halfplanintersection(line *l,int n,Point *poly)
{
sort(l,l+n);
int first,last;
Point *p=new Point[n];
line *q=new line[n];
q[first=last=]=l[];
for(int i=; i<n; i++)
{
while(first<last && !onleft(l[i],p[last-]))last--;
while(first<last && !onleft(l[i],p[first]))first++;
q[++last]=l[i];
if(fabs(cross(q[last].v,q[last-].v))<eps)
{
last--;
if(onleft(q[last],l[i].p))q[last]=l[i];
}
if(first<last)p[last-]=getintersection(q[last-],q[last]);
}
while(first<last && !onleft(q[first],p[last-]))last--;
if((last-first )<=)return ;
p[last]=getintersection(q[last],q[first]);
int m=;
for(int i=first; i<=last; i++)poly[m++]=p[i];
return m;
} Point p[],poly[];
line l[];
Point v[],v2[];
int main()
{
int n,m;
double x,y;
while(scanf("%d",&n)&&n)
{
for(int i=; i<n; i++)
{
scanf("%lf%lf",&x,&y);
p[i]=Point(x,y);
}
for(int i=; i<n; i++)
{
v[i]=p[(i+)%n]-p[i];
v2[i]=nomal(v[i]);
}
double left=0.0,right=20000.0;
while(right-left>eps)
{
double mid=(left+right)/;
for(int i=; i<n; i++)
l[i]=line(p[i]+v2[i]*mid,v[i]);
m=halfplanintersection(l,n,poly);
if(!m)right=mid;
else left=mid;
}
printf("%.6lf\n",left);
}
return ;
}
uva 1396 - Most Distant Point from the Sea的更多相关文章
- 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 ...
- 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 ...
- 1396 - Most Distant Point from the Sea
点击打开链接 题意: 按顺序给出一小岛(多边形)的点 求岛上某点离海最远的距离 解法: 不断的收缩多边形(求半平面交) 直到无限小 二分收缩的距离即可 如图 //大白p263 #include < ...
- 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 ...
- 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 ...
随机推荐
- Android(java)学习笔记209:采用get请求提交数据到服务器(qq登录案例)
1.GET请求: 组拼url的路径,把提交的数据拼装url的后面,提交给服务器. 缺点:(1)安全性(Android下提交数据组拼隐藏在代码中,不存在安全问题) (2)长度有限不能超过4K(h ...
- Flex学习教程网站地址
http://www.985school.com/flex/complex_controls.html
- iOS开发UI篇-tableView在编辑状态下的批量操作(多选)
先看下效果图 直接上代码 #import "MyController.h" @interface MyController () { UIButton *button; } @pr ...
- apache性能配置优化
最近在进行apache性能优化设置.在修改apache配置文件之前需要备份原有的配置文件夹conf,这是网站架设的好习惯.以下的apache配置调优均是在red had的环境下进行的. httpd相关 ...
- Magento 的程序架构与流程
以下是分别详细解读分析程序的各层次源码: MAGENTO_ROOT: //入口文件 /index.php -----–| 1.判断php版本是否大于5.22.引入Magento主要的中心类/app/M ...
- dom+bom
一.判断最大值和最小值,注:arr为数组 最大值:Math.max.apply(null, arr); 最小值:Math.min.apply(null, arr); 二.BOM 打开新页面和关闭打 ...
- 3dmax使用K帧工具创建一个行走动作
第一步,创建一个biped骨骼. 这个就不截图了,因为基本都会. 记住一点,先改变了关键点再去修改,不然修改的是前一个关键帧的动作. 第二步,开启自动关键帧,然后给初始位置双脚添加踩踏关键点.设置关键 ...
- java ssh框架入门
源码:http://pan.baidu.com/s/1hspUOKG
- jQuery Validate验证框架使用
jQuery Validate使用前的准备,需要下载相应js包括:1.jquery.validate.min.js.2.additional-methods.min.js. 当然必不可少的js jQu ...
- C#里面比较时间大小三种方法
1.比较时间大小的实验 string st1="12:13";string st2="14:14";DateTime dt1=Convert.ToDateTim ...