【BZOJ-2618】凸多边形计算几何 + 半平面交 + 增量法 + 三角剖分

2618: [Cqoi2006]凸多边形

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 959 Solved: 489
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Description

逆时针给出n个凸多边形的顶点坐标，求它们交的面积。例如n=2时，两个凸多边形如下图：

则相交部分的面积为5.233。

Input

第一行有一个整数n，表示凸多边形的个数，以下依次描述各个多边形。第i个多边形的第一行包含一个整数m_i，表示多边形的边数，以下m_i行每行两个整数，逆时针给出各个顶点的坐标。

Output

输出文件仅包含一个实数，表示相交部分的面积，保留三位小数。

Sample Input

2
6
-2 0
-1 -2
1 -2
2 0
1 2
-1 2
4
0 -3
1 -1
2 2
-1 0

Sample Output

5.233

HINT

100%的数据满足：2<=n<=10，3<=m_i<=50，每维坐标为[-1000,1000]内的整数

Source

Solution

裸半平面交

这里用的O（n^{2})的增量法，求完半平面交后再求多边形面积就好了，三角剖分一下就可以

一个不错的讲解

具体的做法：

•初始化时加上一个范围巨大的“框”

•每次拿一个新的半平面切割原先的凸集

•保留在新加直线左边的点，删除右边的，有向直线与多边形相交产生的新的点加入到新多边形内

Code

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cmath>

#include<vector>

#include<cstdlib>

using namespace std;

struct Vector

{

    double x,y;

    Vector(double X=,double Y=) {x=X,y=Y;}

};

typedef Vector Point;

typedef vector<Point> Polygon;

Polygon polygon;

#define MAXN 20

#define MAXM 100

Point P[MAXN][MAXM];

#define eps 1e-8

#define INF 1000

const double pi= acos(-1.0);

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});}

int dcmp(double x) {if(fabs(x)<eps) return ; else return x<? -:;}

bool operator == (const Vector& a,const Vector& 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;}

double Len(Vector A) {return sqrt(Dot(A,A));}

double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}

Point GLI(Point P,Vector v,Point Q,Vector w) {Vector u=P-Q; double t=Cross(w,u)/Cross(v,w); return P+v*t;}

double DisTL(Point P,Point A,Point B) {Vector v1=B-A,v2=P-A; return fabs(Cross(v1,v2)/Len(v1));}

bool OnSegment(Point P,Point A,Point B) {return dcmp(DisTL(P,A,B))==&&dcmp(Dot(P-A,P-B))<&&!(P==A)&&!(P==B);}

double PolygonArea(Polygon p)

{

    double area=;

    int n=p.size();

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

        area+=Cross(p[i]-p[],p[i+]-p[]);

    return area/;

}

Polygon CutPolygon(Polygon poly,Point A,Point B)

{

    Polygon newpoly;

    Point C,D,ip;

    int n=poly.size(),i;

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

    {

        C=poly[i];

        D=poly[(i+)%n];

        if(dcmp(Cross(B-A,C-A))>=)

            newpoly.push_back(C);

        if(dcmp(Cross(B-A,D-C))!=)

        {

            ip=GLI(A,B-A,C,D-C);

            if(OnSegment(ip,C,D))

                newpoly.push_back(ip);

        }

    }

    return newpoly;

}

void InitPolygon(Polygon &poly,double inf)

{

    poly.clear();

    poly.push_back((Point){-inf,-inf});

    poly.push_back((Point){inf,-inf});

    poly.push_back((Point){inf,inf});

    poly.push_back((Point){-inf,inf});

}

void Debug()

{

    for (int j=; j<polygon.size(); j++)

        printf("(%.1lf,%.1lf)-->",polygon[j].x,polygon[j].y);

    printf("(%.1lf,%.1lf)",polygon[].x,polygon[].y);

    puts("");

}

int main()

{

    int N,M;

      scanf("%d",&N);

    InitPolygon(polygon,INF);

    for (int i=; i<=N; i++)

        {

            scanf("%d",&M);

            for (int j=; j<=M; j++)

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

            P[i][M+]=P[i][];

            for (int j=; j<=M; j++)

                {polygon=CutPolygon(polygon,P[i][j],P[i][j+]); /*Debug();*/}

        }

    printf("%.3lf\n",PolygonArea(polygon));

    return ;

}

自己调个模板，p事怎么这么多QAQ