zoukankan      html  css  js  c++  java
  • LA 3263 好看的一笔画 欧拉几何+计算几何模板

    题意:训练指南260

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <cmath>
    using namespace std;
    
    struct Point {
        double x, y;
        Point(double x = 0, double y = 0) : 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 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;
    }
    
    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); }
    
    Vector Rotate(Vector A, double rad) {
        return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad) );
    }
    
    Vector Normal(Vector A) {
        double L = Length(A);
        return Vector(-A.y/L, A.x/L);
    }
    
    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;
    }
    
    double DistanceToLine(Point P, Point A, Point B) {
        Vector v1 = B-A, v2 = P - A;
        return fabs(Cross(v1,v2) / Length(v1));
    }
    
    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(v3);
        else return fabs(Cross(v1, v2)) / Length(v1);
    }
    
    Point GetLineProjection(Point P, Point A, Point B) {
        Vector v = B - A;
        return A + v * ( Dot(v, P-A) / Dot(v, v) );
    }
    
    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;
    }
    
    bool OnSegment(Point p, Point a1, Point a2) {
        return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
    }
    
    double ConvexPolygonArea(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;
    }
    
    const int maxn = 300 + 10;
    Point P[maxn], V[maxn*maxn];
    
    Point p[305],v[305*305];
    int main()
    {
        int n,cas=0;
        while(~scanf("%d",&n)&&n)
        {
            cas++;
            for(int i=1;i<=n;i++)
               {
                    scanf("%lf %lf",&p[i].x,&p[i].y);
                    v[i]=p[i];
               }
            n--;
    
            int cnt=n+1;
            for(int i=1;i<=n;i++)
              for(int j=i+1;j<=n;j++)
               if(SegmentProperIntersection(p[i],p[i+1],p[j],p[j+1]))
                    v[++cnt]=GetLineIntersection(p[i],p[i+1]-p[i],p[j],p[j+1]-p[j]);
    
    
            sort(v+1,v+cnt+1);
            int vnum=unique(v+1,v+cnt+1)-(v+1);
            //for(int i=1;i<=vnum;i++)
              //  printf("v %d:%f %f
    ",i,v[i].x,v[i].y);
            int e=n;
            //cout<<"ori   "<<e<<endl;
            for(int i=1;i<=vnum;i++)
                for(int j=1;j<=n;j++)
                  if(OnSegment(v[i],p[j],p[j+1]))
                      e++;
           //cout<<"vnum:"<<vnum<<"  e:"<<e<<endl;
            printf("Case %d: There are %d pieces.
    ",cas,e+2-vnum);
        }
        return 0;//v+f-e=2;
    }
    

      

  • 相关阅读:
    [Luogu P3626] [APIO2009] 会议中心
    杭电 1869 六度分离 (求每两个节点间的距离)
    杭电 1874 畅通工程续 (求某节点到某节点的最短路径)
    最短路径模板
    杭电 2544 最短路径
    POJ 1287 Networking (最小生成树模板题)
    NYOJ 1875 畅通工程再续 (无节点间距离求最小生成树)
    POJ 2485 Highways (求最小生成树中最大的边)
    杭电 1233 还是畅通工程 (最小生成树)
    杭电 1863 畅通工程 (最小生成树)
  • 原文地址:https://www.cnblogs.com/smilesundream/p/5267183.html
Copyright © 2011-2022 走看看