裸的半平面交求多边形是否有核.
多边形的核: 在多边形核上的点可以看到多边形的所有顶点,凸多边形的核显然就是多边形本身.
多边形的核是一个凸包,对多边形的所有边都做向着多边形的半平面交在判断一下是否构成凸包就可以了
一样的题目还有POJ3335
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Video Surveillance
Description A friend of yours has taken the job of security officer at the Star-Buy Company, a famous depart- ment store. One of his tasks is to install a video surveillance system to guarantee the security of the customers (and the security of the merchandise of course) The first problem is to choose where to install the camera for every floor. The only requirement is that every part of the room must be visible from there. In the following figure the left floor can be completely surveyed from the position indicated by a dot,
Before trying to install the cameras, your friend first wants to know whether there is indeed a suitable position for them. He therefore asks you to write a program that, given a ground plan, de- termines whether there is a position from which the whole floor Input The input contains several floor descriptions. Every description starts with the number n of vertices that bound the floor (4 <= n <= 100). The next n lines contain two integers each, the x and y coordinates for the n vertices, given in clockwise order. All A zero value for n indicates the end of the input. Output For every test case first output a line with the number of the floor, as shown in the sample output. Then print a line stating "Surveillance is possible." if there exists a position from which the entire floor can be observed, or print "Surveillance is impossible." Print a blank line after each test case. Sample Input 4 0 0 0 1 1 1 1 0 8 0 0 0 2 1 2 1 1 2 1 2 2 3 2 3 0 0 Sample Output Floor #1 Surveillance is possible. Floor #2 Surveillance is impossible. Source |
/* ***********************************************
Author :CKboss
Created Time :2015年04月09日 星期四 19时43分00秒
File Name :POJ1474.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define mp make_pair
#define pb push_back
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = 1e20;
const int maxp = 111;
int dcmp(double d) { if(fabs(d)<eps) return 0; return (d<0)?-1:1; }
inline double sqr(double x) { return x*x; }
struct point
{
double x,y;
point(double _x=0,double _y=0):x(_x),y(_y){}
void input() { scanf("%lf%lf",&x,&y); }
void output() { printf("%.2lf %.2lf\n",x,y); }
bool operator==(point a) const { return dcmp(a.x-x)==0&&dcmp(a.y-y)==0; }
bool operator<(point a) const { return dcmp(a.x-x)==0?dcmp(y-a.y)<0:x<a.x; }
double len() { return hypot(x,y); }
double len2() { return x*x+y*y; }
double distance(point p) { return hypot(x-p.x,y-p.y); }
point add(point p) { return point(x+p.x,y+p.y); }
point sub(point p) { return point(x-p.x,y-p.y); }
point mul(double b) { return point(x*b,y*b); }
point div(double b) { return point(x/b,y/b); }
double dot(point p) { return x*p.x+y*p.y; }
double det(point p) { return x*p.y-y*p.x; }
double rad(point a,point b)
{
point p=*this;
return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
}
};
struct line
{
point a,b;
line(){}
line(point _a,point _b) { a=_a; b=_b; }
bool operator==(const line v) const { return (a==v.a)&&(b==v.b); }
bool parallel(line v) { return dcmp(b.sub(a).det(v.b.sub(v.a)))==0; }
point crosspoint(line v)
{
double a1=v.b.sub(v.a).det(a.sub(v.a));
double a2=v.b.sub(v.a).det(b.sub(v.a));
return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1));
}
};
struct halfplane:public line
{
double angle;
halfplane(){}
/// a-->b left
halfplane(point _a,point _b){ a=_a; b=_b;}
halfplane(line v) { a=v.a; b=v.b; }
void calcangle() { angle=atan2(b.y-a.y,b.x-a.x); }
bool operator<(const halfplane &b) const { return angle<b.angle; }
};
struct halfplanes
{
int n;
halfplane hp[maxp];
point p[maxp];
int que[maxp];
int st,ed;
void push(halfplane tmp)
{
hp[n++]=tmp;
}
void unique()
{
int m=1,i;
for(i=1;i<n;i++)
{
if(dcmp(hp[i].angle-hp[i-1].angle)) hp[m++]=hp[i];
else if(dcmp(hp[m-1].b.sub(hp[m-1].a).det(hp[i].a.sub(hp[m-1].a))>0))
hp[m-1]=hp[i];
}
n=m;
}
bool halfplaneinsert()
{
int i;
for(int i=0;i<n;i++) hp[i].calcangle();
sort(hp,hp+n);
unique();
que[st=0]=0; que[ed=1]=1;
p[1]=hp[0].crosspoint(hp[1]);
for(i=2;i<n;i++)
{
while(st<ed&&dcmp((hp[i].b.sub(hp[i].a).det(p[ed].sub(hp[i].a))))<0) ed--;
while(st<ed&&dcmp((hp[i].b.sub(hp[i].a).det(p[st+1].sub(hp[i].a))))<0) st++;
que[++ed]=i;
if(hp[i].parallel(hp[que[ed-1]])) return false;
p[ed]=hp[i].crosspoint(hp[que[ed-1]]);
}
while(st<ed&&dcmp((hp[st].b.sub(hp[que[st]].a).det(p[ed].sub(hp[que[st]].a))))<0) ed--;
while(st<ed&&dcmp((hp[que[ed]].b.sub(hp[que[ed]].a).det(p[st+1].sub(hp[que[ed]].a))))<0) st++;
if(st+1>=ed) return false;
return true;
}
};
int n;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int cas=1;
while(scanf("%d",&n)&&n)
{
vector<point> vp;
for(int i=0;i<n;i++)
{
point tp; tp.input(); vp.pb(tp);
}
halfplanes hfs; hfs.n=0;
for(int i=0;i<n;i++)
{
//// p[i] <--- p[i+1]
hfs.push(halfplane(vp[(i+1)%n],vp[i]));
}
printf("Floor #%d\n",cas++);
bool isok=hfs.halfplaneinsert();
if(isok==true) puts("Surveillance is possible.");
else puts("Surveillance is impossible.");
putchar(10);
}
return 0;
}