Luogu-4196 [CQOI2006]凸多边形

凸多边形的面积就相当于半平面交

求几个凸多边形面积交就相当于一堆半平面一起求交

#include<cmath>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn=1e3+100;

struct Point{

    double x,y;

    Point(double xx=0,double yy=0){

        x=xx,y=yy;

    }

}a[maxn],p[maxn];

struct Vector{

    double x,y;

    Vector(double xx=0,double yy=0){

        x=xx,y=yy;

    }

};

struct Line{

    Point p;

    Vector v;

    double ang;

    Line(Point a=Point(),Vector b=Vector()){

        p=a,v=b;

        ang=atan2(v.y,v.x);

    }

}q[maxn],b[maxn];

int dcmp(double x){return fabs(x)<1e-9?0:(x>0?1:-1);}

Vector operator - (Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}

Point operator + (Point a,Vector b){return Point(a.x+b.x,a.y+b.y);}

Vector operator * (double p,Vector a){return Vector(a.x*p,a.y*p);}

double operator * (Vector a,Vector b){return a.x*b.y-a.y*b.x;}

double operator * (Point a,Point b){return a.x*b.y-a.y*b.x;}

bool operator < (Line x,Line y){return dcmp(x.ang-y.ang)==0?(dcmp(x.v*(y.p-x.p))>0):(x.ang<y.ang);}

Point glt(Line x,Line y){Vector v=x.p-y.p; return x.p+(y.v*v)/(x.v*y.v)*x.v;}

bool onright(Line a,Line b,Line t){Point p=glt(a,b); return dcmp(t.v*(p-t.p))<0;}

void bpm(Line *b,int &n,Point *p){

    sort(b+1,b+n+1);

    int l=0,r=1,tot=0;

    for(int i=1;i<=n;i++){

        if(b[i].ang!=b[i-1].ang) tot++;

        b[tot]=b[i];

    }

    n=tot,q[0]=b[1],q[1]=b[2];

    for(int i=3;i<=n;i++){

        while(l<r&&onright(q[r],q[r-1],b[i])) r--;

        while(l<r&&onright(q[l],q[l+1],b[i])) l++;

        q[++r]=b[i];

    }

    while(l<r&&onright(q[r],q[r-1],q[l])) r--;

    while(l<r&&onright(q[l],q[l+1],q[r])) l++;

    n=0,q[r+1]=q[l];

    for(int i=l;i<=r;i++)

        b[++n]=q[i],p[n]=glt(q[i],q[i+1]);

}

double S(Point *p,int n){

    double ans=0;

    p[n+1]=p[1];

    for(int i=1;i<=n;i++)

        ans+=p[i]*p[i+1];

    return fabs(ans)/2;

}

int t,m,n;

int main(){

    scanf("%d",&t);

    while(t--){

        scanf("%d",&m);

        for(int i=1;i<=m;i++)

            scanf("%lf%lf",&a[i].x,&a[i].y);

        a[m+1]=a[1];

        for(int i=1;i<=m;i++)

            b[++n]=Line(a[i],a[i+1]-a[i]);

    }

    bpm(b,n,p);

    printf("%.3lf\n",S(p,n));

    return 0;

}