[ACM_几何] Fishnet

 

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28417#problem/C

本题大意：有一个1X1的矩形，每边按照从小到大的顺序给n个点如图，然后对应连线将举行划分，求最大面积。

解题思路：暴力算出各点，求出各面积

#include<iostream>
#include<cmath>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<iomanip>

using namespace std;
#define eps 0.0000000001
#define PI acos(-1.0)

//点和向量
struct Point{
    double x,y;
    Point(double x=,double y=):x(x),y(y){}
};
typedef Point Vector;
Vector operator+(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator-(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator*(Vector a,double p){return Vector(a.x*p,a.y*p);}
Vector operator/(Vector a,double p){return Vector(a.x/p,a.y/p);}
bool operator<(const Vector& a,const Vector& b){return a.x<b.x||(a.x==b.x && a.y<b.y);}
int dcmp(double x){
    if(fabs(x)<eps)return ;
    else return x< ? -:;
}
bool operator==(const Point& a,const Point& b){
    return dcmp(a.x-b.x)== && dcmp(a.y-b.y)==;
}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//向量点积
double Length(Vector A){return sqrt(Dot(A,A));}//向量模长
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}//向量夹角
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);}//三角形面积的2倍
//绕起点逆时针旋转rad度
Vector Rotate(Vector A,double rad){
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
double torad(double jiao){return jiao/*PI;}//角度转弧度
double tojiao(double ang){return ang/PI*;}//弧度转角度
//单位法向量
Vector Normal(Vector A){
    double L=Length(A);
    return Vector(-A.y/L,A.x/L);
}
//点和直线
struct Line{
    Point P;//直线上任意一点
    Vector v;//方向向量，他的左边对应的就是半平面
    double ang;//极角，即从x正半轴旋转到向量v所需的角（弧度）
    Line(){}
    Line(Point p,Vector v):P(p),v(v){ang=atan2(v.y,v.x);}
    bool operator<(const Line& L)const {
        return ang<L.ang;
    }
};
//计算直线P+tv和Q+tw的交点（计算前必须确保有唯一交点）即：Cross(v,w)非0
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
//点到直线距离(dis between point P and line AB)
double DistanceToLine(Point P,Point A,Point B){
    Vector v1=B-A , v2=P-A;
    return fabs(Cross(v1,v2))/Length(v1);
}
//dis between point P and segment AB
double DistancetoSegment(Point P,Point A,Point B){
    if(A==B)return Length(P-A);
    Vector v1=B-A,v2=P-A,v3=P-B;
    if(dcmp(Dot(v1,v2))<)return  Length(v2);
    else if(dcmp(Dot(v1,v3))>)return Length(v3);
    else return fabs(Cross(v1,v2))/Length(v1);
}
//point P on line AB 投影点
Point GetLineProjection(Point P,Point A,Point B){
    Vector v=B-A;
    return A+v*(Dot(v,P-A)/Dot(v,v));
}
//线段规范相交（只有一个且不在端点）每条线段两端都在另一条两侧，（叉积符号不同）
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){
    double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
           c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)< && dcmp(c3)*dcmp(c4)<;
}
//判断点P是否在线段AB上
bool OnSegment(Point p,Point a1,Point a2){
    return dcmp(Cross(a1-p,a2-p))== && dcmp(Dot(a1-p,a2-p))<;
}
//多边形的面积（可以是非凸多边形）
double PolygonArea(Point* p,int n){
    double area=;
    for(int i=;i<n-;i++)
        area+=Cross(p[i]-p[],p[i+]-p[]);
    return area/;
}

//点p在有向直线左边，上面不算
bool OnLeft(Line L,Point p){
    return Cross(L.v,p-L.P)>;
}

double ok(double x,double y,double d,double z){
    double f=fabs(d*(/tan(acos(z/y))+/tan(acos(z/x))))-z;
    if(fabs(f)<1e-)return  ;
    else return f;
}
//计算凸包输入点数组p，个数n，输出点数组ch，返回凸包定点数
//输入不能有重复，完成后输入点顺序被破坏
//如果不希望凸包的边上有输入点，把两个<=改成<
//精度要求高时，建议用dcmp比较
//基于水平的Andrew算法-->1、点排序2、删除重复的然后把前两个放进凸包
//3、从第三个往后当新点在凸包前进左边时继续，否则一次删除最近加入的点，直到新点在左边
int ConVexHull(Point* p,int n,Point*ch){
    sort(p,p+n);
    int m=;
    for(int i=;i<n;i++){//下凸包
        while(m> && Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=)m--;
        ch[m++]=p[i];
    }
    int k=m;
    for(int i=n-;i>=;i--){//上凸包
        while(m>k && Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=)m--;
        ch[m++]=p[i];
    }
    if(n>)m--;
    return m;
}
int main(){
    Point point[][];
    Point a[],b[],c[],d[];
    for(int n;cin>>n&&n;){

        point[][].x=;
        point[][].y=;
        for(int i=;i<n;i++){
            cin>>a[i].x;
            a[i].y=;
            point[][i+]=a[i];
        }

        point[n+][n+].x=;
        point[n+][n+].y=;
        for(int i=;i<n;i++){
            cin>>b[i].x;
            b[i].y=;
            point[n+][i+]=b[i];
        }

        point[n+][].x=;
        point[n+][].y=;
        for(int i=;i<n;i++){
            cin>>c[i].y;
            c[i].x=;
            point[i+][]=c[i];
        }

        point[][n+].x=;
        point[][n+].y=;
        for(int i=;i<n;i++){
            cin>>d[i].y;
            d[i].x=;
            point[i+][n+]=d[i];
        }

        for(int i=;i<n;i++){
            for(int j=;j<n;j++){
                point[i+][j+]=GetLineIntersection(c[i],d[i]-c[i],a[j],b[j]-a[j]);
            }
        }

        Point four[];
        double max=-,area;
        for(int i=;i<=n;i++){
            for(int j=;j<=n+;j++){
                four[]=point[i][j];
                four[]=point[i][j+];
                four[]=point[i+][j+];
                four[]=point[i+][j];
                area=PolygonArea(four,);
                if(area>max)max=area;
            }
        }
        cout<<fixed<<max<<'\n';
    }return ;
}