• HDU 4667 Building Fence 计算几何 凸包+圆


    1.三角形的所有端点

    2.过所有三角形的端点对所有圆做切线,得到所有切点。

    3.做任意两圆的外公切线,得到所有切点。

    对上述所有点求凸包,标记每个点是三角形上的点还是某个圆上的点。

    求完凸包后,因为所有点都是按逆时针(或顺时针)排好序的,如果相邻两点在同一圆上,那么求这段圆弧的距离,否则求这段直线的距离。最后得到所有周长。

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    
    using namespace std;
    
    const double eps = 1e-9;
    const double PI = acos(-1.0);
    const int MAXN = 60;
    
    struct Point
    {
        double x, y;
        int id;         //点标号,标记是否在同一个圆上
        Point() { }
        Point( double x, double y ):x(x), y(y) { }
        Point( double x, double y, int id ):x(x), y(y), id(id) { }
        void readPoint()
        {
            scanf( "%lf%lf", &x, &y );
            return;
        }
    };
    
    struct Circle
    {
        Point c;   //圆心坐标
        double r;  //半径
        Circle() {}
        Circle( Point c, double r ): c(c), r(r) {}
        Point getPoint( double theta )   //根据极角返回圆上一点的坐标
        {
            return Point( c.x + cos(theta)*r, c.y + sin(theta)*r );
        }
        void readCircle()
        {
            scanf("%lf%lf%lf", &c.x, &c.y, &r );
            return;
        }
    };
    
    typedef Point Vector;
    
    Vector operator+( Vector A, Vector B )       //向量加
    {
        return Vector( A.x + B.x, A.y + B.y );
    }
    
    Vector operator-( Vector A, Vector B )       //向量减
    {
        return Vector( A.x - B.x, A.y - B.y );
    }
    
    Vector operator*( Vector A, double p )      //向量数乘
    {
        return Vector( A.x * p, A.y * p );
    }
    
    Vector operator/( Vector A, double p )      //向量数除
    {
        return Vector( A.x / p, A.y / p );
    }
    
    int dcmp( double x )    //控制精度
    {
        if ( fabs(x) < eps ) return 0;
        else return x < 0 ? -1 : 1;
    }
    
    bool operator<( const Point& A, const Point& B )   //两点比较
    {
        return dcmp( A.x - B.x) < 0 || ( dcmp(A.x - B.x ) == 0 && dcmp( A.y - B.y ) < 0 );
    }
    
    bool operator>( const Point& A, const Point& B )   //两点比较
    {
        return dcmp( A.x - B.x) > 0 || ( dcmp(A.x - B.x ) == 0 && dcmp( A.y - B.y ) > 0 );
    }
    
    bool operator==( const Point& a, const Point& b )   //两点相等
    {
        return dcmp( a.x - b.x ) == 0 && dcmp( a.y - b.y ) == 0;
    }
    
    double Cross( Vector A, Vector B )   //向量叉积
    {
        return A.x * B.y - A.y * B.x;
    }
    
    double PointDis( Point a, Point b ) //两点距离的平方
    {
        return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y);
    }
    
    //求凸包,graham算法,O(nlogn),返回凸包点的个数
    int graham( Point *p, int n, Point *ch )
    {
        if ( n <= 2 ) return 0;
        int top = 0;
        sort( p, p + n );
    
        ch[ top ] = p[0];
        ch[ ++top ] = p[1];
        ch[ ++top ] = p[2];
    
        top = 1;
    
        for ( int i = 2; i < n; ++i )
        {
            while ( top && dcmp( Cross( ch[top] - ch[top - 1], p[i] - ch[top - 1] ) ) <= 0 ) --top;
            ch[++top] = p[i];
        }
        int len = top;
        ch[++top] = p[n - 2];
        for ( int i = n - 3; i >= 0; --i )
        {
            while ( top > len && dcmp( Cross( ch[top] - ch[top - 1], p[i] - ch[top - 1] ) ) <= 0 ) --top;
            ch[++top] = p[i];
        }
        return top;
    }
    
    //过定点做圆的切线,得到切点,返回切点个数
    //tps保存切点坐标
    int getTangentPoints( Point p, Circle C, Point *tps )
    {
        int cnt = 0;
    
        double dis = sqrt( PointDis( p, C.c ) );
        int aa = dcmp( dis - C.r );
        if ( aa < 0 ) return 0;  //点在圆内
        else if ( aa == 0 ) //点在圆上,该点就是切点
        {
            tps[cnt] = p;
            ++cnt;
            return cnt;
        }
    
        //点在圆外,有两个切点
        double base = atan2( p.y - C.c.y, p.x - C.c.x );
        double ang = acos( C.r / dis );
        //printf( "base = %f ang=%f
    ", base, ang );
        //printf( "base-ang=%f  base+ang=%f 
    ", base - ang, base + ang );
    
        tps[cnt] = C.getPoint( base - ang ), ++cnt;
        tps[cnt] = C.getPoint( base + ang ), ++cnt;
    
        return cnt;
    }
    
    //求两圆外公切线切点,返回切线个数
    //p是圆c2在圆c1上的切点
    int makeCircle( Circle c1, Circle c2, Point *p )
    {
        int cnt = 0;
        double d = sqrt( PointDis(c1.c, c2.c) ), dr = c1.r - c2.r;
        double b = acos(dr / d);
        double a = atan2( c2.c.y - c1.c.y, c2.c.x - c1.c.x );
        double a1 = a - b, a2 = a + b;
        p[cnt++] = Point(cos(a1) * c1.r, sin(a1) * c1.r) + c1.c;
        p[cnt++] = Point(cos(a2) * c1.r, sin(a2) * c1.r) + c1.c;
        return cnt;
    }
    
    double DisOnCircle( Point a, Point b, Circle c )  //求圆上一段弧长
    {
        Point o = c.c;
        double A = sqrt( PointDis( o, a ) );
        double B = sqrt( PointDis( o, b ) );
        double C = sqrt( PointDis( a, b ) );
        double alpha = acos( ( A*A + B*B - C*C ) / ( 2.0*A*B ) );
        if ( dcmp( Cross( o-a, o-b ) ) < 0 ) return alpha * c.r;
        else return ( 2.0*PI - alpha ) * c.r;
    }
    
    /**********************以上模板**********************/
    
    int cntC, cntT;  //圆的个数,三角形的个数
    Circle yuan[MAXN];   //所有圆
    Point PP[300100];    //所有点
    Point tubao[300100]; //凸包
    int totPP;           //点总数
    
    void showP( Point *p, int nn )
    {
        printf( "allP = %d
    ", nn );
        for ( int i = 0; i < nn; ++i )
            printf("%f %f
    ", p[i].x, p[i].y );
        puts("-------------------------");
        return;
    }
    
    int main()
    {
        //freopen( "10022.in", "r", stdin );
        //freopen( "s.out", "w", stdout );
        while ( scanf( "%d%d", &cntC, &cntT ) == 2 )
        {
            totPP = 0;
            for ( int i = 0; i < cntC; ++i )
                yuan[i].readCircle();
            for ( int i = 0; i < cntT; ++i )
            {
                for ( int j = 0; j < 3; ++j )
                {
                    PP[totPP].readPoint();
                    PP[totPP].id = -(totPP+2);
                    ++totPP;
                }
            }
    
            if ( cntC == 1 && cntT == 0 )
            {
                printf("%.6f
    ", 2.0 * PI * yuan[0].r );
                continue;
            }
    
            int pretot = totPP;
            //求两圆的外切点
            for ( int i = 0; i < cntC; ++i )
                for ( int j = i + 1; j < cntC; ++j )
                {
                    Point PonA[4], PonB[4];
                    makeCircle( yuan[i], yuan[j], PonA );
                    int ans = makeCircle( yuan[j], yuan[i], PonB );
                    for ( int k = 0; k < ans; ++k )
                    {
                        PonA[k].id = i;
                        PonB[k].id = j;
                        PP[totPP++] = PonA[k];
                        PP[totPP++] = PonB[k];
                    }
                }
    
            //求所有点与所有圆的切点
            for ( int i = 0; i < pretot; ++i )
            {
                for ( int j = 0; j < cntC; ++j )
                {
                    Point qiedian[4];
                    int ans = getTangentPoints( PP[i], yuan[j], qiedian );
                    for ( int k = 0; k < ans; ++k )
                    {
                        qiedian[k].id = j;
                        PP[totPP++] = qiedian[k];
                    }
                }
            }
    
            //showP( PP, totPP );
            int cntBao = graham( PP, totPP, tubao );
            //puts("*********");
            //showP( tubao, cntBao );
            double girth = 0.0;
            tubao[cntBao] = tubao[0];
    
            for ( int i = 1; i <= cntBao; ++i )
            {
                if ( tubao[i].id == tubao[i - 1].id )  //如果两点在同一个圆上
                    girth += DisOnCircle( tubao[i], tubao[i - 1], yuan[ tubao[i].id ] );
                else
                    girth += sqrt( PointDis( tubao[i], tubao[i - 1] ) );
    
            }
    
            printf( "%.5lf
    ", girth );
        }
        return 0;
    }
  • 相关阅读:
    在 json4s 中自定义CustomSerializer
    【重点】2020年宝山区义务教育阶段学校校区范围与招生计划(小学)
    2019宝山区小升初对口地段表及对口初中片区划分
    2019上海市各区重点幼儿园、小学和中学排名(建议收藏)
    转:一千行MYSQL 笔记
    基于weixin-java-mp 做微信JS签名 invalid signature签名错误 官方说明
    转 : 深入解析Java锁机制
    微服务架构转型升级
    抽奖活动 mark
    抽奖 mark
  • 原文地址:https://www.cnblogs.com/GBRgbr/p/3260917.html
Copyright © 2020-2023  润新知