[bzoj4445] [SCOI2015]小凸想跑步 (半平面交)

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题意:凸包上一个点\(p\),使得\(p\)和点\(0,1\)组成的三角形面积最小
用叉积来求:
\(p,i,i+1\)组成的三角形面积为: (\(\times\)为叉积)
\((p_p-i)\times (p_p-p_{i+1})\Rightarrow\)
\((x_p-x_i,y_p-y_i)\times(x_p-x_{i+1},y_p-y_{i+1})\Rightarrow\)
\((x_p-x_i)(y_p-y_{i+1})-(y_p-y_i)(x_p-x_{i+1})\Rightarrow\)
\(x_py_p-x_py_{i+1}-x_iy_p+x_iy_{i+1}-x_py_p+x_{i+1}y_p+x_py_i-x_{i+1}y_i\Rightarrow\)
\(x_p(y_i-y_{i+1})+y_p(x_{i+1}-x_i)+(x_iy_{i+1}-x_{i+1}y_i)\)
要求点\(p\)和点\(0,1\)组成的三角形面积最小,即:
\(x_p(y_0-y_1)+y_p(x_1-x_0)+(x_0y_1-x_1y_0)<x_p(y_i-y_{i+1})+y_p(x_{i+1}-x_i)+(x_iy_{i+1}-x_{i+1}y_i)\Rightarrow\)
\(x_p(y_0-y_1-y_i+y_{i+1})+y_p(x_1-x_0-x_{i+1}+x_i)+(x_0y_1-x_1y_0-x_iy_{i+1}+x_{i+1}y_i)<0\)
可以发现,方程为\(ax+by+c<0\)的形式,可以求出\(n\)个方程,和原凸多边形求一下半平面交,交出来的面积与原多边形面积的比值即为答案

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<bitset>
#include<sstream>
#include<cstdlib>
#define QAQ int
#define TAT long long
#define OwO bool
#define ORZ double
#define Ug unsigned
#define F(i,j,n) for(QAQ i=j;i<=n;++i)
#define E(i,j,n) for(QAQ i=j;i>=n;--i)
#define MES(i,j) memset(i,j,sizeof(i))
#define MEC(i,j) memcpy(i,j,sizeof(j))

using namespace std;
const QAQ N=300005;
const ORZ eps=1e-8;
QAQ sign(ORZ x){return fabs(x)<=eps ? 0 : (x>0 ? 1 : -1);}

QAQ n;
struct Point {
    ORZ x,y;
    Point(){}
    Point(ORZ X,ORZ Y){x=X;y=Y;}
    
    friend Point operator + (Point a,Point b){
        return Point(a.x+b.x,a.y+b.y);
    }
    friend Point operator - (Point a,Point b){
        return Point(a.x-b.x,a.y-b.y);
    }
    friend Point operator * (Point a,ORZ k){
        return Point(a.x*k,a.y*k);
    }
    friend ORZ operator * (Point a,Point b){
        return a.x*b.x+a.y*b.y;
    }
    friend ORZ operator ^ (Point a,Point b){
        return a.x*b.y-a.y*b.x;
    }
}p[N];
struct Line{
    Point p,v;
    ORZ poa;
    
    Line(){}
    Line(Point a,Point b){
        p=a;v=b;
        poa=atan2(b.y,b.x);
    }
    friend OwO operator < (Line a,Line b){
        return sign(a.poa-b.poa)==0 ? sign((a.v) ^ (b.p-a.p)) >0 : sign(a.poa-b.poa)<0;
    }
}a[N],q[N];
QAQ js,head,tail,cnt;
ORZ s1,s2;

Point inter(Line a,Line b){
    Point u=a.p-b.p;
    ORZ k=(b.v^u)/(a.v^b.v);
    return a.p+a.v*k;
}

OwO pd(Line a,Point b){
    return sign(a.v^(b-a.p))>=0;
}

void Half_Plane(){
    sort(a+1,a+js+1);
    cnt=1;
    F(i,2,js) if(sign(a[i].poa-a[cnt].poa)>0) a[++cnt]=a[i];
    head=1;tail=0;
    q[++tail]=a[1];q[++tail]=a[2];
    F(i,3,cnt){
        while(head<tail&&pd(a[i],inter(q[tail-1],q[tail]))) tail--;
        while(head<tail&&pd(a[i],inter(q[head+1],q[head]))) head++;
        q[++tail]=a[i];
    }
    while(head<tail&&pd(q[head],inter(q[tail-1],q[tail]))) tail--;
    F(i,head,tail-1) p[i]=inter(q[i],q[i+1]);
    p[tail]=inter(q[tail],q[head]);
    F(i,head,tail-1) s2+=(p[i]^(p[i+1]-p[i]));
    s2+=(p[tail]^(p[head]-p[tail]));
}

QAQ main(){
    scanf("%d",&n);
    F(i,0,n-1) scanf("%lf%lf",&p[i].x,&p[i].y);
    p[n]=p[0];
    F(i,0,n-1) {
        a[++js]=Line(p[i+1],p[i]-p[i+1]);
        s1+=(p[i]^(p[i+1]-p[i]));
    }
    F(i,1,n-1){
        ORZ A=p[i+1].x-p[i].x-p[1].x+p[0].x;
        ORZ B=p[i+1].y-p[i].y-p[1].y+p[0].y;
        ORZ C=-(p[i]^(p[i+1]-p[i]))+(p[0]^(p[1]-p[0]));
        if(sign(A)!=0) a[++js]=Line(Point(0,C/A),Point(-A,-B));
        else if(sign(B)!=0) a[++js]=Line(Point(-C/B,0),Point(0,-B));
    }
    Half_Plane();
    printf("%.4lf\n",fabs(s2/s1));
    return 0;
}

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