(beginer) 半平面交 UVA 588 - Video Surveillance
2014-02-18 14:20
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Video Surveillance |
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 file contains several floor descriptions. Every description starts with the numbern of vertices that bound the floor (
). 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 ``Surveillanceis 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.
Miguel A. Revilla
1998-03-10
题意:在一个所有边都是平行于坐标轴的多边形中是否有一点放了一个摄像头之后能看到这个多边形的任意一个地方。
思路:这道题和之前的一个题目有点像,不过那个题目是求可能位置的总面积。其实差不多的。这道题目就是看半平面是不是空集。这里面形成的半平面可能是一个直线、一个点这类的,这种情况也是可以的。所以之前的半平面的算法要改一改。把OnLeft的改成点在线上面也返回true,这就过了。
代码:
#include<iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<string>
#include<set>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define mp make_pair
#define eps 1e-10
const double inf = 1e16;
const double PI = acos(-1.0);
struct Point
{
Point(const Point&p) { x = p.x , y = p.y; }
Point (double xx=0,double yy=0) : x(xx) , y(yy) { }
double x;
double y;
};
typedef Point Vector;
Vector operator+(Vector v1,Vector v2) { return Vector(v1.x+v2.x,v1.y+v2.y); }
Vector operator-(Vector v1,Vector v2) { return Vector(v1.x-v2.x,v1.y-v2.y); }
Vector operator*(Vector v, double p) { return Vector(v.x*p,v.y*p); }
Vector operator/(Vector v,double p) { return Vector(v.x/p,v.y/p); }
int dcmp(double x)
{
if (fabs(x) < eps) return 0;
return x < 0 ? -1 : 1;
}
bool operator < (Point a,Point b) { return dcmp(a.x-b.x)<0 || dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)<0; }
bool operator==(const Point & a,const Point & b)
{
return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
inline double toRad(double x) { return x * PI/180; }
inline double toDegreed(double rad) { return rad*180/PI; }
double Dot(Vector A,Vector B) { return A.x*B.x+A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A,A)); }
double Angle(Vector A,Vector B)
{
double cosine = Dot(A,B)/Length(A)/Length(B);
if (dcmp(cosine-1.0)==0) return 0;
return acos(cosine);
}
double Cross(Vector A,Vector B) { return A.x*B.y-A.y*B.x; }
double Area2(Point a,Point b,Point c) { return Cross(b-a,c-a); }
double PolyArea(Point *poly, int n)
{
double ret = 0;
for (int i = 1 ; i < n-1 ; ++i)
ret += Cross(poly[i]-poly[0],poly[i+1]-poly[0]);
return ret/2;
}
//旋转
Vector Rotate(Vector A,double rad)
{
return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
//单位法线
Vector Normal(Vector A) { double L = Length(A); return Vector(-A.y/L,A.x/L); }
//点和直线
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
{
Vector u = P-Q;
double t = Cross(w,u) / Cross(v,w);
return P+v*t;
}
double DistanceToLine(Point P,Point A,Point B)
{
Vector v1 = B-A , v2 = P-A;
return fabs(Cross(v1,v2))/Length(v1);
}
double DistanceToSegment(Point P,Point A,Point B)
{
if (A==B) return Length(P-A);
Vector v1 = B-A , v2 = P-A , v3 = P-B;
if (dcmp(Dot(v1,v2)) < 0) return Length(v2);
else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);
else return fabs(Cross(v1,v2))/Length(v1);
}
//点在直线上的投影
Point GetLineProjection(Point P,Point A,Point B)
{
Vector v = B-A;
return A+v*(Dot(v,P-A)/Dot(v,v));
}
//线段相交(不包括端点)
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
double c1 = Cross(a2-a1,b1-a1) , c2 = Cross(a2-a1,b2-a1) ,
c3 = Cross(b2-b1,a1-b1) , c4 = Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4)<0;
}
//点在线段上(不包括端点)
bool OnSegment(Point p,Point a,Point b)
{
return dcmp(Cross(a-p,b-p))==0 && dcmp(Dot(a-p,b-p)) < 0;
}
//点在线上
bool OnLine(Point p,Point a,Point b)
{
if (p==a || p==b) return true;
return dcmp(Cross(a-p,b-p))==0;
}
//线段相交(包括端点)
bool SegmentIntersection(Point a1,Point a2,Point b1,Point b2)
{
if (SegmentProperIntersection(a1,a2,b1,b2)) return true;
if (OnSegment(a1,b1,b2) || OnSegment(a2,b1,b2)) return true;
if (OnSegment(b1,a1,a2) || OnSegment(b2,a1,a2)) return true;
if (a1==b1 || a1==b2 || a2==b1 || a2==b2) return true;
return false;
}
//点在多边形内:0:外部 1:内部 -1:线上 -2:在顶点
int isPointInPolygon(Point a,Point *p,int n)
{
int wn = 0;
for (int i = 0 ; i < n ; ++i) {
if (a==p[i]) return -2; //点重合
if (OnSegment(a,p[i],p[(i+1)%n])) return -1;
int k = dcmp(Cross(p[(i+1)%n]-p[i],a-p[i]));
int d1 = dcmp(p[i].y-a.y);
int d2 = dcmp(p[(i+1)%n].y-a.y);
if (k > 0 && d1<=0 && d2>0) ++wn;
if (k<0 && d2<=0 && d1>0) --wn;
}
if (wn!=0) return 1; //内部
return 0; //外部
}
//--------------------------------------------------------------------------------------------
//直线和直线
struct Line
{
Point P; //直线上任意一点
Vector v; // 方向向量。它的左边就是对应的半平面
double ang; //极角
Line() { }
Line(Point P,Vector v)
{
this->P = P ; this->v = v;
ang = atan2(v.y,v.x);
}
bool operator < (const Line& L) const { return ang < L.ang; } //排序用的比较运算符
Point point(double t) { return v*t+P; }
};
//点p在有向直线L的左边(线上不算)
bool OnLeft(Line L , Point p) { return Cross(L.v,p-L.P) > 0; }
//二直线交点。假定交点唯一存在
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;
}
//--------------------------------------------
//与圆相关
struct Circle
{
Circle() { }
Point c;
double r;
Circle(Point c, double r) : c(c) , r(r) { }
Point point (double a) { return Point(c.x+cos(a)*r,c.y+sin(a)*r); }
};
int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point>& sol)
{
double a = L.v.x , b = L.P.x-C.c.x , c= L.v.y, d = L.P.y-C.c.y;
double e = a*a+c*c , f = 2*(a*b+c*d) , g = b*b+d*d-C.r*C.r;
double delta = f*f-4*e*g; //判别式
if (dcmp(delta) < 0) return 0; //相离
if (dcmp(delta)==0) { //相切
t1 = t2 = -f/(2*e);
sol.push_back(L.point(t1));
return 1;
}
//相交
t1 = (-f-sqrt(delta)) / (2*e); sol.push_back(L.point(t1));
t2 = (-f+sqrt(delta)) / (2*e); sol.push_back(L.point(t2));
return 2;
}
double angle(Vector v) { return atan2(v.y,v.x); }
int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol)
{
double d = Length(C1.c-C2.c);
if (dcmp(d)==0) {
if (dcmp(C1.r-C2.r)==0) return -1; //两圆重合
return 0;
}
if (dcmp(C1.r+C2.r-d) < 0) return 0;
if (dcmp(fabs(C1.r-C2.r)-d) > 0) return 0;
double a = angle(C2.c-C1.c);
double da = acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); //向量C1C2的极角
//C1C2到C1P1的角
Point p1 = C1.point(a-da) , p2 = C1.point(a+da);
sol.push_back(p1);
if (p1==p2) return 1;
sol.push_back(p2);
return 2;
}
//国电p到圆C的切线。v[i]是第i条切线的向量。返回切线条数
int getTangents(Point p,Circle C,Vector* v)
{
Vector u= C.c-p;
double dist = Length(u);
if (dist < C.r) return 0;
else if (dcmp(dist-C.r)==0) {
v[0] = Rotate(u,PI/2);
return 1;
} else {
double ang = asin(C.r/dist);
v[0] = Rotate(u,-ang);
v[1] = Rotate(u,+ang);
return 2;
}
}
int getTangents(Circle A,Circle B,Point* a, Point* b)
{
int cnt = 0;
if (A.r < B.r) { swap(A,B); swap(a,b); }
double d2 = Dot(A.c-B.c,A.c-B.c);
double rdiff = A.r-B.r;
double rsum = A.r+B.r;
if (dcmp(d2-rdiff*rdiff) < 0) return 0; //内含
double base = atan2(B.c.y-A.c.y,B.c.x-A.c.x);
if (d2==0 && dcmp(A.r-B.r)==0) return -1; //无限多条切线
if (dcmp(d2-rdiff*rdiff)==0) { //内切,1条切线
a[cnt] = A.point(base);
b[cnt] = B.point(base);
++cnt;
return 1;
}
//有外切共线
double ang = acos((A.r-B.r)/sqrt(d2));
a[cnt] = A.point(base+ang); b[cnt] = B.point(base+ang); ++cnt;
a[cnt] = A.point(base-ang); b[cnt] = B.point(base-ang); ++cnt;
if (dcmp(d2-rsum*rsum)==0) { //一条内公切线
a[cnt] = A.point(base);
b[cnt] = B.point(PI+base);
++cnt;
} else if (dcmp(d2-rsum*rsum)>0) { //两条公切线
double ang = acos((A.r+B.r)/sqrt(d2));
a[cnt] = A.point(base+ang); b[cnt] = B.point(PI+base+ang); ++cnt;
a[cnt] = A.point(base-ang); b[cnt] = B.point(PI+base-ang); ++cnt;
}
return cnt;
}
//点p和圆的关系: 0:在圆上 1:在圆外 -1:在圆内
int PointCircleRelation(Point p,Circle c)
{
return dcmp(Dot(p-c.c,p-c.c)-c.r*c.r);
}
//A在B内
bool InCircle(Circle A,Circle B)
{
if (dcmp(A.r-B.r)>0) return false;
double d2 = Dot(A.c-B.c,A.c-B.c);
double rdiff = A.r-B.r;
double rsum = A.r+B.r;
if (dcmp(d2-rdiff*rdiff) <= 0) return true; //内含或内切或重合
return false;
}
//----------------------------------------------------------------------------
//与球相关的转化
//角度转换为弧度
double torad(double deg) { return deg/180*PI; }
//经纬度(角度)转化为空间坐标
void get_coord(double R,double lat,double lng,double& x,double& y,double& z)
{
lat = torad(lat);
lng = torad(lng);
x = R*cos(lat)*cos(lng);
y = R*cos(lat)*sin(lng);
z = R*sin(lat);
}
//-----------------------------------------------------------------------
//几何算法:
//凸包:O(nlogn) 得到的凸包逆时针
int ConvexHull(Point* p , int n, Point* ch)
{
sort(p,p+n);
int m = 0;
for (int i = 0 ; i < n ; ++i) {
while (m>1 && dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<=0) --m;
ch[m++] = p[i];
}
int k = m;
for (int i = n-2 ; i >= 0 ; --i) {
while (m>k && dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<=0) --m;
ch[m++] = p[i];
}
if (n > 1) --m;
return m;
}
//半平面交:(OnLeft>改成=变成最后的半平面可以是一条直线)
int HalfplaneIntersection(Line* L,int n,Point* poly)
{
sort(L,L+n); //按极角排序
int first , last; //双端队列的第一个元素和最后一个元素的下表
Point *p = new Point
; //p[i]为q[i]和q[i+1]的交点
Line *q = new Line
; //双端队列
q[first=last=0] = L[0]; //双端队列初始化为只有一个半平面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;
q[++last] = L[i];
if (dcmp(Cross(q[last].v,q[last-1].v))==0) {
//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] = GetIntersection(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 ) { delete [] p; delete [] q; return 0; } //空集(**)
p[last] = GetIntersection(q[last],q[first]); //计算首尾两个半平面的交点
//从deque复制到输出中
int m = 0;
for (int i = first ; i <= last ; ++i) poly[m++] = p[i];
delete [] p; delete[] q;
return m ;
}
bool OnLine(Point p,Line L)
{
Vector v = p-L.P;
return dcmp(Cross(v,L.v))==0;
}
//旋转卡壳:
pair<double,double> RotateCalipers(Point* hull,int n)
{
hull
= hull[0];
Line L1, L2, L3, L4;
int q=0, w=0, e=0, r=0;
for (int i = 0 ; i < n ; ++i) {
if (dcmp(hull[i].y-hull[q].y)<0) q = i;
if (dcmp(hull[i].x-hull[w].x)<0) w = i;
if (dcmp(hull[i].y-hull[e].y)>0) e = i;
if (dcmp(hull[i].x-hull[r].x)>0) r = i;
}
L1 = Line(hull[q],Vector(1,0));
L2 = Line(hull[w],Vector(0,-1));
L3 = Line(hull[e],Vector(-1,0));
L4 = Line(hull[r],Vector(0,1));
double sum = 0;
double ansA = inf , ansC = inf;
while (dcmp(sum-PI/2)<0) {
double ang = Angle(hull[q+1]-hull[q],L1.v);
ang = min(ang,Angle(hull[w+1]-hull[w],L2.v));
ang = min(ang,Angle(hull[e+1]-hull[e],L3.v));
ang = min(ang,Angle(hull[r+1]-hull[r],L4.v));
sum += ang;
L1.v = Rotate(L1.v,ang);
L2.v = Rotate(L2.v,ang);
L3.v = Rotate(L3.v,ang);
L4.v = Rotate(L4.v,ang);
if (dcmp(ang-Angle(hull[q+1]-hull[q],L1.v))==0) { L1.P = hull[q+1]; q = (q+1)%n; }
if (dcmp(ang-Angle(hull[w+1]-hull[w],L2.v))==0) { L2.P = hull[w+1]; w = (w+1)%n; }
if (dcmp(ang-Angle(hull[e+1]-hull[e],L3.v))==0) { L3.P = hull[e+1]; e = (e+1)%n; }
if (dcmp(ang-Angle(hull[r+1]-hull[r],L4.v))==0) { L4.P = hull[r+1]; r = (r+1)%n; }
double A = DistanceToLine(L1.P,L3.P,L3.P+L3.v*Length(L1.P-L3.P));
double B = DistanceToLine(L2.P,L4.P,L4.P+L4.v*Length(L2.P-L4.P));
ansA = min(ansA,A*B);
ansC = min(ansC,A+B);
}
ansC *= 2;
return mp(ansA,ansC);
}
//-----------------------------------------------------------------------
const int maxn = 100+10;
int n;
Point p[maxn];
Line L[maxn];
void input()
{
for (int i = 0 ; i < n ; ++i) scanf("%lf%lf",&p[i].x,&p[i].y);
p
= p[0];
for (int i = 0 ; i < n; ++i) L[i] = Line(p[i+1],p[i]-p[i+1]);
}
void solve()
{
int m = HalfplaneIntersection(L,n,p);
if (m==0) puts("Surveillance is impossible.");
else puts("Surveillance is possible.");
}
int main()
{
int k = 0;
while (~scanf("%d",&n) && n) {
input();
++k;
printf("Floor #%d\n",k);
solve();
printf("\n");
}
}
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