poj3335 Rotating Scoreboard
题目描述:
题解:
半平面交判核的存在性。
重点在于一个点的核也算核。
这样的话普通的求多边形的版本就要加一个特判。
就是把剩下的一个节点暴力带回所有直线重判,这时判叉积是否$\leq 0$,而不是$<0$。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<algorithm> using namespace std; const int N = 150; const double eps = 1e-8; int dcmp(double x) { if(fabs(x)<=eps)return 0; return x>0?1:-1; } struct Point { double x,y; Point(){} Point(double x,double y):x(x),y(y){} Point operator + (const Point&a)const{return Point(x+a.x,y+a.y);} Point operator - (const Point&a)const{return Point(x-a.x,y-a.y);} Point operator * (const double&a)const{return Point(x*a,y*a);} double operator ^ (const Point&a)const{return x*a.y-y*a.x;} }; typedef Point Vector; typedef vector<Point> Pol; double ang(const Vector&v){return atan2(v.x,v.y);} struct Line { Point p; Vector v; Line(){} Line(Point p,Vector v):p(p),v(v){} bool operator < (const Line&a)const { return ang(v)<ang(a.v); } }; bool Onleft(Line a,Point p) { return dcmp((a.v^(p-a.p)))>0; } bool onleft(Line a,Point p) { return dcmp((a.v^(p-a.p)))>=0; } Point L_L(Line a,Line b) { double t = ((b.p-a.p)^b.v)/(a.v^b.v); return a.p+a.v*t; } int T,n; Point p[N],tp[N]; Line s[N],ts[N]; int bpmj() { memset(ts,0,sizeof(ts));memset(tp,0,sizeof(tp)); sort(s+1,s+1+n); int hd,tl; ts[hd=tl=1]=s[1]; for(int i=2;i<=n;i++) { while(hd<tl&&!Onleft(s[i],tp[tl-1]))tl--; while(hd<tl&&!Onleft(s[i],tp[hd]))hd++; ts[++tl] = s[i]; if(!dcmp(ts[tl].v^ts[tl-1].v)) { tl--; if(Onleft(ts[tl],s[i].p))ts[tl]=s[i]; } if(hd<tl)tp[tl-1] = L_L(ts[tl-1],ts[tl]); } while(hd<tl&&!Onleft(ts[hd],tp[tl-1]))tl--; if(hd==tl) { Point now = tp[hd]; for(int i=1;i<=n;i++) if(!onleft(s[i],now)) return 0; } return 1; } void build1() { for(int i=1;i<n;i++)s[i]=Line(p[i],p[i+1]-p[i]); s[n] = Line(p[n],p[1]-p[n]); } void build2() { for(int i=2,j=n;i<j;i++,j--)swap(p[i],p[j]); for(int i=1;i<n;i++)s[i]=Line(p[i],p[i+1]-p[i]); s[n] = Line(p[n],p[1]-p[n]); } void work() { scanf("%d",&n);int ans = 0; for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); build1(); ans|=bpmj(); build2(); ans|=bpmj(); if(ans)puts("YES"); else puts("NO"); } int main() { // freopen("tt.in","r",stdin); scanf("%d",&T); while(T--)work(); return 0; }