半平面交的具体过程原理再次不再过多赘述，只是整理一些细节。
1、设置边界，为了防止边界出锅，我们需要框一个大框，具体长这样。
b[++tot]=node(-inf,inf,-inf,-inf);b[++tot]=node(-inf,-inf,inf,-inf);
b[++tot]=node(inf,-inf,inf,inf);b[++tot]=node(inf,inf,-inf,inf);
2、把每条线段构造出来，然后用atan2算出线段与x轴的夹角。
for(int i=1;i<=tot;++i)b[i].ang=atan2(b[i].y.y-b[i].x.y,b[i].y.x-b[i].x.x);
3、将所有线段按照夹角排序，如果夹角相等就看哪条线段更靠里，这个可以用叉积算。
bool operator <(const node &b)const{ if(fabs(ang-b.ang)<eps)return (y-x)*(b.x-x)<eps; else return ang<b.ang; }
4、这个非常关键，如果有平行的线段在后面算的时候会出锅，所以我们要去重，因为我们排过序了，所以我们直接保留相同夹角的第一条线段就好了。
for(int i=1;i<=tot;++i)if(fabs(b[i].ang-b[i-1].ang)>eps)c[++top]=b[i]; for(int i=1;i<=top;++i)b[i]=c[i];tot=top;
5、加入每条线段，维护单调队列。
h=1;t=2;q[h]=b[1];q[t]=b[2];p[h]=jiao(b[1],b[2]); for(int i=3;i<=tot;++i){ while(h<t&&left(b[i],p[t-1]))t--; while(h<t&&left(b[i],p[h]))h++; q[++t]=b[i];p[t-1]=jiao(q[t],q[t-1]); } while(h<t&&left(q[t],p[h]))h++; while(h<t&&left(q[h],p[t-1]))t--;
6、维护时我们要判断一条线段是否在点的左边，这个可以用叉积算。
inline bool left(node b,P a){return (a-b.x)*(b.y-b.x)>eps;}
7、我们开需要求出两条线段的交点，这个用面积(还是叉积)算。
inline point jiao(line a,line b){return b.a+(b.b-b.a)*(((b.a-a.a)*(a.b-a.a))/((a.b-a.a)*(b.b-b.a)));}
8、最后我们用叉积算出多边形的面积
for(int i=h+2;i<=t;++i)ans+=(p[i]-p[h])*(p[i-1]-p[h])/2;
HNOI2003多边形
一开始以为要先极角排序一下，然后WA到死。
发现点是按顺序给的。
#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #define double double #define N 2002 #define inf 2e9 using namespace std; int n,top,tot,h,t; double ans; const double eps=1e-9; struct point{ double x,y; 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.y-y*b.x;} point operator *(const double &b)const{return point{x*b,y*b};} }d[N],p[N]; inline double dis(point a,point b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);} struct line{ point a,b;double ang; line(){} line(double aa,double bb,double cc,double dd){a.x=aa;a.y=bb;b.x=cc;b.y=dd;} bool operator <(const line &x)const{ if(fabs(ang-x.ang)<eps)return (b-a)*(x.a-a)<eps; else return ang<x.ang; } }l[N],c[N],q[N]; inline bool cmp(point a,point b){ double cj=(a-d[1])*(b-d[1]); return (cj>0)||(fabs(cj)<eps&&dis(a,d[1])<dis(b,d[1])); } inline bool left(line a,point b){return (b-a.a)*(a.b-a.a)>eps;} inline point jiao(line a,line b){return b.a+(b.b-b.a)*(((b.a-a.a)*(a.b-a.a))/((a.b-a.a)*(b.b-b.a)));} int main(){ scanf("%d",&n); for(int i=1;i<=n;++i)scanf("%lf%lf",&d[i].x,&d[i].y); /* int ji=1; for(int i=2;i<=n;++i)if(d[i].x<d[ji].x||(d[i].x==d[ji].x&&d[i].y<d[ji].y))ji=i; swap(d[ji],d[1]); sort(d+2,d+n+1,cmp);*/ reverse(d+1,d+n+1); for(int i=2;i<=n;++i)l[i-1]=line(d[i-1].x,d[i-1].y,d[i].x,d[i].y); l[n]=line(d[n].x,d[n].y,d[1].x,d[1].y);tot=n; l[++tot]=line(-inf,inf,-inf,-inf);l[++tot]=line(-inf,-inf,inf,-inf); l[++tot]=line(inf,-inf,inf,inf);l[++tot]=line(inf,inf,-inf,inf); for(int i=1;i<=tot;++i)l[i].ang=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x); sort(l+1,l+tot+1); for(int i=1;i<=tot;++i)if(fabs(l[i].ang-l[i-1].ang)>eps)c[++top]=l[i]; for(int i=1;i<=top;++i)l[i]=c[i];tot=top; h=1;t=2;q[h]=l[1];q[t]=l[2];p[1]=jiao(q[1],q[2]); for(int i=3;i<=tot;++i){ while(h<t&&left(l[i],p[t-1]))t--; while(h<t&&left(l[i],p[h]))h++; q[++t]=l[i];p[t-1]=jiao(q[t],q[t-1]); } while(h<t&&left(q[t],p[h]))h++; while(h<t&&left(q[h],p[t-1]))t--;// p[t]=jiao(q[h],q[t]); for(int i=h+2;i<=t;++i)ans+=(p[i]-p[h])*(p[i-1]-p[h])/2; ans=fabs(ans); printf("%.2lf",ans); return 0; }
CQOI2006凸多边形
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #define N 10002 #define inf 2e9 #define double long double using namespace std; const double eps=1e-9; int n,tot,h,t,m,top; double x[N],y[N],ans; inline int rd(){ int x=0;char c=getchar();bool f=0; while(!isdigit(c)){if(c=='-')f=1;c=getchar();} while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();} return f?-x:x; } struct P{ double x,y; P operator +(const P &b)const{return P{x+b.x,y+b.y};} P operator -(const P &b)const{return P{x-b.x,y-b.y};} double operator *(const P &b)const{return x*b.y-y*b.x;} P operator *(const double &b)const{return P{x*b,y*b};} double operator ^(const P &b)const{return x*b.x+y*b.y;} inline void print(){ cout<<x<<" "<<y<<endl; } }p[N]; struct node{ P x,y;double ang; node(){} node(int a,int b,int c,int d){x.x=a;x.y=b;y.x=c;y.y=d;ang=0;} bool operator <(const node &b)const{ if(fabs(ang-b.ang)<eps)return (y-x)*(b.x-x)<eps; else return ang<b.ang; } inline void print(){ cout<<x.x<<" "<<x.y<<" "<<y.x<<" "<<y.y<<endl; } }b[N],q[N],c[N]; inline P jiao(node a,node b){ double k=((b.x-a.x)*(a.y-a.x))/((a.y-a.x)*(b.y-b.x)); return b.x+(b.y-b.x)*k; } inline bool left(node b,P a){return (a-b.x)*(b.y-b.x)>eps;} int main(){ n=rd(); b[++tot]=node(-inf,inf,-inf,-inf);b[++tot]=node(-inf,-inf,inf,-inf); b[++tot]=node(inf,-inf,inf,inf);b[++tot]=node(inf,inf,-inf,inf); for(int i=1;i<=n;++i){ m=rd(); x[1]=rd();y[1]=rd(); for(int j=2;j<=m;++j){ x[j]=rd();y[j]=rd(); b[++tot]=node(x[j-1],y[j-1],x[j],y[j]); } b[++tot]=node(x[m],y[m],x[1],y[1]); } for(int i=1;i<=tot;++i)b[i].ang=atan2(b[i].y.y-b[i].x.y,b[i].y.x-b[i].x.x); sort(b+1,b+tot+1); for(int i=1;i<=tot;++i)if(fabs(b[i].ang-b[i-1].ang)>eps)c[++top]=b[i]; for(int i=1;i<=top;++i)b[i]=c[i];tot=top; h=1;t=2;q[h]=b[1];q[t]=b[2];p[h]=jiao(b[1],b[2]); for(int i=1;i<=tot;++i){ while(h<t&&left(b[i],p[t-1]))t--; while(h<t&&left(b[i],p[h]))h++; q[++t]=b[i];p[t-1]=jiao(q[t],q[t-1]); } while(h<t&&left(q[t],p[h]))h++; while(h<t&&left(q[h],p[t-1]))t--; if(t-h+1<=2){printf("0.000 ");return 0;} p[t]=jiao(q[h],q[t]); for(int i=h+2;i<=t;++i)ans+=(p[i]-p[h])*(p[i-1]-p[h])/2; ans=fabs(ans); printf("%.3LF",ans); return 0; }
HNOI2012射箭
瞎**化简一下二次函数，发现这就是半平面交，外面套个二分答案就可以了。
#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #define inf 1e14 #define N 220002 #define double long double using namespace std; typedef long long ll; int tot,n; const double eps=1e-14; inline ll rd(){ ll x=0;char c=getchar();bool f=0; while(!isdigit(c)){if(c=='-')f=1;c=getchar();} while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();} return f?-x:x; } struct node{double x,y1,y2;}; struct point{ double x,y; 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.y-y*b.x;} point operator *(const double &b)const{return point{x*b,y*b};} }p[N]; struct line{ point a,b;double ang;int id; line(){id=0;} line(double x1,double x2,double x3,double x4){a.x=x1;a.y=x2;b.x=x3;b.y=x4;id=ang=0;} bool operator <(const line &x)const{ if(ang==x.ang)return (b-a)*(x.a-a)>eps; return ang<x.ang; } }l[N],a[N],q[N]; inline point jiao(line a,line b){//????? double k=((a.b-a.a)*(b.a-a.a))/((b.b-b.a)*(a.b-a.a)); return b.a+(b.b-b.a)*k; } bool left(point x,line b){return (x-b.a)*(b.b-b.a)>eps;} inline bool check(int mid){ int top=0; for(int i=1;i<=tot;++i)if(l[i].id<=mid)if(!top||fabs(l[i].ang-a[top].ang)>eps)a[++top]=l[i]; int h=1,t=2;q[1]=a[1];q[2]=a[2];p[1]=jiao(a[1],a[2]); for(int i=3;i<=top;++i){ while(h<t&&left(p[t-1],a[i]))t--; while(h<t&&left(p[h],a[i]))h++; q[++t]=a[i];p[t-1]=jiao(q[t-1],q[t]); } while(h<t&&left(p[t-1],q[h]))t--; while(h<t&&left(p[h],q[t]))h++; //cout<<mid<<" "<<h<<" "<<t<<endl; return t-h>=2; } int main(){ n=rd();double x,y1,y2; l[++tot]=line(-inf,inf,-inf,-inf);l[++tot]=line(-inf,-inf,inf,-inf); l[++tot]=line(inf,-inf,inf,inf);l[++tot]=line(inf,inf,-inf,inf); for(int i=1;i<=n;++i){ x=rd();y1=rd();y2=rd(); l[++tot].a.x=-1;l[tot].a.y=y1/x+x; l[tot].b.x=1;l[tot].b.y=y1/x-x;l[tot].id=i; l[++tot].a.x=1;l[tot].a.y=y2/x-x; l[tot].b.x=-1;l[tot].b.y=y2/x+x;l[tot].id=i; } for(int i=1;i<=tot;++i)l[i].ang=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x); sort(l+1,l+tot+1); int l=0,r=n,ans=0; while(l<=r){ int mid=(l+r)>>1; if(check(mid)){ans=mid;l=mid+1;} else r=mid-1; } printf("%d",ans); return 0; }