Video Surveillance
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 3145 | Accepted: 1391 |
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) on all of the store‘s countless floors. As the company has only a limited budget, there will be only one camera on every floor. But these cameras may turn around to look in every direction.
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,
while for the right floor, there is no such position, the given position failing to see the lower left part of the floor.
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
is visible. All floor ground plans form rectangular polygons, whose edges do not intersect each other and touch each other only at the corners.
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 vertices will be distinct and at corners of the polygon. Thus the edges alternate between horizontal and vertical.
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." if there is no such position.
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.
给定一个多边形，求多边形的核。
将多边形每一条边延长，然后计算半平面交。
判断点与直线关系由于模板少了一个等于号，导致当半平面退化成单点的时候判断出错，虽然浪费了一天的时间，但是改正了模板的一个错误，
还是值得的。
代码:
/* ***********************************************
Author :_rabbit
Created Time :2014/5/4 15:03:55
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 10000000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x>0?1:-1;
}
struct Point{
double x,y;
Point(double _x=0,double _y=0){
x=_x;y=_y;
}
};
Point operator + (const Point &a,const Point &b){
return Point(a.x+b.x,a.y+b.y);
}
Point operator - (const Point &a,const Point &b){
return Point(a.x-b.x,a.y-b.y);
}
Point operator * (const Point &a,const double &p){
return Point(a.x*p,a.y*p);
}
Point operator / (const Point &a,const double &p){
return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
return a.x*b.x+a.y*b.y;
}
double Length(Point a){
return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
return a/Length(a);
}
Point Normal(Point a){
return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
return Length(Cross(b-a,c-a));
}
struct Line{
Point p,v;
double ang;
Line(){};
Line(Point p,Point 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 dcmp(Cross(L.v,p-L.p))>=0;
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
Point u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
Point GetLineIntersection(Line a,Line b){
return GetLineIntersection(a.p,a.v,b.p,b.v);
}
vector<Point> HPI(vector<Line> L){
int n=L.size();
sort(L.begin(),L.end());//将所有半平面按照极角排序。
/*for(int i=0;i<n;i++){
cout<<"han "<<i<<" ";
printf("%.2lf %.2lf %.2lf %.2lf %.2lf\n",L[i].p.x,L[i].p.y,L[i].v.x,L[i].v.y,L[i].ang);
}*/
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--;//删除顶部的半平面
while(first<last&&!OnLeft(L[i],p[first]))first++;//删除底部的半平面
q[++last]=L[i];//将当前的半平面假如双端队列顶部。
if(fabs(Cross(q[last].v,q[last-1].v))<eps){//对于极角相同的，选择性保留一个。
last--;
if(OnLeft(q[last],L[i].p))q[last]=L[i];
}
if(first<last)p[last-1]=GetLineIntersection(q[last-1],q[last]);//计算队列顶部半平面交点。
}
while(first<last&&!OnLeft(q[first],p[last-1]))last--;//删除队列顶部的无用半平面。
if(last-first<=1)return ans;//半平面退化
p[last]=GetLineIntersection(q[last],q[first]);//计算队列顶部与首部的交点。
for(int i=first;i<=last;i++)ans.push_back(p[i]);//将队列中的点复制。
return ans;
}
double PolyArea(vector<Point> p){
int n=p.size();
double ans=0;
for(int i=1;i<n-1;i++)
ans+=Cross(p[i]-p[0],p[i+1]-p[0]);
return fabs(ans)/2;
}
Point pp[200];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n;
int cc=0;
while(~scanf("%d",&n)&&n){
for(int i=0;i<n;i++)scanf("%lf%lf",&pp[i].x,&pp[i].y);pp[n]=pp[0];
Point a,b;
vector<Line> L;
Line s;
a=Point(-INF,-INF);b=Point(INF,-INF);s=Line(a,b-a);L.push_back(s);
a=Point(INF,-INF);b=Point(INF,INF);s=Line(a,b-a);L.push_back(s);
a=Point(INF,INF);b=Point(-INF,INF);s=Line(a,b-a);L.push_back(s);
a=Point(-INF,INF);b=Point(-INF,-INF);s=Line(a,b-a);L.push_back(s);
for(int i=0;i<n;i++){
L.push_back(Line(pp[i],pp[i]-pp[i+1]));
}
vector<Point> ans=HPI(L);
if (cc) cout << endl;cout << "Floor #" << ++cc << endl;
if (!ans.size()) cout << "Surveillance is impossible.\n";
else cout << "Surveillance is possible.\n";
}
return 0;
}
POJ 1474 多边形的核(半平面交）