题目链接:https://vjudge.net/problem/POJ-1584
题意:首先要判断凸包,然后判断圆是否在多边形中。
思路:
判断凸包利用叉积,判断圆在多边形首先要判断圆心是否在多边形中,然后判断圆心到每条边的距离是否小于半径。板子很重要!!
AC code:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std; const double eps=1e-;
const double inf=1e20;int sgn(double x){
if(abs(x)<eps) return ;
if(x<) return -;
return ;
} struct Point{
double x,y;
Point(){}
Point(double xx,double yy):x(xx),y(yy){}
Point operator + (const Point& b)const{
return Point(x+b.x,y+b.y);
}
Point operator - (const Point& b)const{
return Point(x-b.x,y-b.y);
}
double operator * (const Point& b)const{
return x*b.x+y*b.y;
}
double operator ^ (const Point& b)const{
return x*b.y-b.x*y;
}
//绕原点旋转角度b(弧度值),后x、y的变化
void transXY(double b){
double tx=x,ty=y;
x=tx*cos(b)-ty*sin(b);
y=tx*sin(b)+ty*cos(b);
}
}; struct Line{
Point s,e;
Line(){}
Line(Point ss,Point ee){
s=ss,e=ee;
}
//两直线相交求交点
//第一个值为0表示直线重合,为1表示平行,为2表示相交
//只有第一个值为2时,交点才有意义
pair<int,Point> operator &(const Line &b)const{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == )
{
if(sgn((s-b.e)^(b.s-b.e)) == )
return make_pair(,res);//重合
else return make_pair(,res);//平行
}
double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
res.x += (e.x-s.x)*t;
res.y += (e.y-s.y)*t;
return make_pair(,res);
}
};
//判断线段相交
bool inter(Line l1,Line l2){
return
max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x)&&
max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x)&&
max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y)&&
max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y)&&
sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=&&
sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s))<=;
} double dis(Point a,Point b){
return sqrt((b-a)*(b-a));
}
//判断点在线段上
bool OnSeg(Point P,Line L){
return
sgn((L.s-P)^(L.e-P))==&&
sgn((P.x-L.s.x)*(P.x-L.e.x))<=&&
sgn((P.y-L.s.y)*(P.y-L.e.y))<=;
}
//判断点在凸多边形内,复杂度O(n)
//点形成一个凸包,而且按逆时针排序(如果是顺时针把里面的<0改为>0)
//点的编号:0~n-1
//返回值:
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inConvexPoly(Point a,Point p[],int n){
for(int i=;i<n;++i)
if(sgn((p[i]-a)^(p[(i+)%n]-a))<) return -;
else if(OnSeg(a,Line(p[i],p[(i+)%n]))) return ;
return ;
}
//判断点在任意多边形内,复杂度O(n)
//射线法,poly[]的顶点数要大于等于3,点的编号0~n-1,按逆时针或顺时针排序
//返回值
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inPoly(Point a,Point p[],int n){
int cnt=;
Line ray,side;
ray.s=a;
ray.e.y=a.y;
ray.e.x=-inf;
for(int i=;i<n;++i){
side.s=p[i];
side.e=p[(i+)%n];
if(OnSeg(a,side)) return ;
if(sgn(side.s.y-side.e.y)==) continue;
if(OnSeg(side.s,ray)){
if(sgn(side.s.y-side.e.y)>) ++cnt;
}
else if(OnSeg(side.e,ray)){
if(sgn(side.e.y-side.s.y)>) ++cnt;
}
else if(inter(ray,side)) ++cnt;
}
if(cnt%==) return ;
else return -;
}
//判断凸多边形
//允许共线边
//点可以是顺时针给出也可以是逆时针给出
//点的编号是0~n-1
bool isconvex(Point poly[],int n){
bool s[];
memset(s,false,sizeof(s));
for(int i=;i<n;++i){
s[sgn((poly[(i+)%n]-poly[i])^(poly[(i+)%n]-poly[i]))+]=true;
if(s[]&&s[]) return false;
}
return true;
}
//点到线段的距离
//返回点到线段最近的点
Point NearestPointToLineSeg(Point P,Line L){
Point result;
double t=((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
if(t>=&&t<=){
result.x=L.s.x+(L.e.x-L.s.x)*t;
result.y=L.s.y+(L.e.y-L.s.y)*t;
}
else{
if(dis(P,L.s)<dis(P,L.e))
return L.s;
else
return L.e;
}
return result;
} int n;
double R,X,Y;
Point pt[]; int main(){
while(scanf("%d",&n),n>=){
scanf("%lf%lf%lf",&R,&X,&Y);
for(int i=;i<n;++i)
scanf("%lf%lf",&pt[i].x,&pt[i].y);
if(!isconvex(pt,n)){
printf("HOLE IS ILL-FORMED\n");
continue;
}
Point P=Point(X,Y);
if(inPoly(P,pt,n)==-){
printf("PEG WILL NOT FIT\n");
continue;
}
int flag=;
for(int i=;i<n;++i){
if(sgn(dis(P,NearestPointToLineSeg(P,Line(pt[i],pt[(i+)%n])))-R)<){
flag=;
break;
}
}
if(flag) printf("PEG WILL FIT\n");
else printf("PEG WILL NOT FIT\n");
}
return ;
}