LA 3263 欧拉定理

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1264

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 310;
const int maxe = 100000;
const int INF = 0x3f3f3f;

struct Point{
    double x,y;
    Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
};
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 Point& a,const Point& b){
    return a.x < b.x ||( a.x == b.x && a.y < b.y);
}
const double eps = 1e-10;
int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else              return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b){
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

///向量（x,y）的极角用atan2(y,x);
double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
double Length(Vector A)    { return sqrt(Dot(A,A)); }
double Angle(Vector A, Vector B)  { return acos(Dot(A,B) / Length(A) / Length(B)); }

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); }

///向量的逆时针旋转，rad 为旋转的角；
Vector Rotate(Vector A, double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); }
///特殊的，下面函数计算向量的单位法向量，即左旋90，在长度归一化；
Vector Normal(Vector A){
    double L = Length(A);
    return Vector(-A.y/L, A.x/L);
}

///直线用参数式 P = P'+ t * v；P'为直线上一点，v为方向向量；
///t不受限制，为直线，t>0 为射线， 0=<t<=1为线段；
///推导暂时不会,下面计算直线P+tv和Q+tw的交点。调用前先确保两直线有唯一交点；即判定Cross(v,w) 非0；
Point GetLineIntersecion(Point P, Vector v,Point Q,Vector w){
    Vector u = P - Q;
    double t = Cross(w,u)/Cross(v,w);
    return P + v*t;
}

///求点到直线的距离，利用h*|AB| == AB(向量) * AP(向量)；
double DistanceToLine(Point P,Point A,Point B){
    Vector v1 = B - A, v2 = P - A;
    return fabs(Cross(v1,v2)) / Length(v1);
}

///求点P到线段AB的距离，先看Q点在线段外还是内；利用点积就可以，
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)) < 0)       return Length(v2);
    else if(dcmp(Dot(v1,v3) > 0))  return Length(v2);
    else    return  fabs(Cross(v1,v2))/Length(v1);
}
///如果要求Q的话：（当然满足Q在线段内）,由公式Dot(v,P-(A+t'*v)) == 0 推出；
Point GetLineProjection(Point P,Point A,Point B){
    Vector v = B - A;
    return A + v * (Dot(v,P-A)/Dot(v,v));
}

///判定线段是否规范相交；
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) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
///如果允许在端点处相交：c1和c2都是0，表示两线段共线；如果只有其中一个为0，则一条线段的端点在另一条线段上；
///下面的代码判断一个点P是否在一条线段AB上(不包括A,B点)；
bool OnSegment(Point P,Point A,Point B){
    return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(P-A,P-B)) < 0;
}

///多边形
///求面积
double PolygonArea(Point* p,int n){
    double area = 0;
    for(int i=1;i<n-1;i++){
        area += Cross(p[i]-p[0],p[i+1]-p[0]);
    }
    return area/2;
}


int main()
{
    //freopen("E:\acm\input.txt","r",stdin);
    //freopen("E:\acm\output.txt","w",stdout);

    Point p[maxn],v[maxn*maxn];
    int n,T=1;
    while(scanf("%d",&n)==1 && n){
       for(int i=0;i<n;i++) { scanf("%lf %lf",&p[i].x,&p[i].y); v[i] = p[i]; }
        n--;
        int vcnt = n, ecnt = n;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++){
               if( SegmentProperIntersection(p[i],p[i+1],p[j],p[j+1]) ){
                  v[vcnt++] = GetLineIntersecion(p[i],p[i+1]-p[i],p[j],p[j+1]-p[j]);
               }
        }
        sort(v,v+vcnt);  ///排序等会好去重；
        vcnt = unique(v,v+vcnt) - v;  ///求出不重复的点的个数；
        
        for(int i=0;i<vcnt;i++)
            for(int j=0;j<n;j++)
              if( OnSegment(v[i],p[j],p[j+1]) )
                  ecnt++;
        printf("Case %d: There are %d pieces.
",T++,ecnt+2-vcnt);
    }

    return 0;
}