HDU 4667 Building Fence 解题报告

比赛

题目

题意：

给n个圆和m个三角形，且保证互不相交，用一个篱笆把他们围起来，求最短的周长是多少。

题解：

解法1：在每个圆上均匀的取2000个点，求凸包周长就可以水过。

解法2：求出所有圆之间的外公切线的切点，以及过三角形每个顶点的的直线和圆的切点，和三角形的三个顶点。这些点做凸包确定篱笆边上的图形。凸包的边和圆弧之和即为所求。求圆弧长度的时候要判断是优弧还是劣弧。用叉积判断两个向量的方向关系即可。

//Time：218MS//Memory：860Kinclude <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>using namespace std;const double EPS = 1e-10;const double PI = acos(-1.0);const int MAXN = 55;int dcmp(double x){    if(fabs(x)<EPS) return 0;    return x<0? -1:1;}double sqr(double x){    return x*x;}struct Point{    double x,y;    bool tp;    int id;    Point(){}    Point(double a,double b):x(a),y(b){}    Point operator +(const Point &a)const{return Point(x+a.x,y+a.y);}    Point operator -(const Point &a)const{return Point(x-a.x,y-a.y);}    Point operator *(double k) const{return Point(x*k,y*k);}    Point operator /(double k) const{return Point(x/k,y/k);}    bool operator <(const Point &a)const    {        return dcmp(x-a.x)<0||(dcmp(x-a.x)==0&&dcmp(y-a.y)<0);    }    bool operator ==(const Point &a)const    {        return dcmp(x-a.x)==0&&dcmp(y-a.y)==0;    }    Point trunc(double d)    {        double dist(Point ,Point);        double len = dist(*this,Point(0,0));        return Point(x*d/len,y*d/len);    }    Point rotate(double a)    {        return Point(x*cos(a)-y*sin(a),y*cos(a)+x*sin(a));    }    void input(){scanf("%lf%lf",&x,&y);}};struct Circle{    Point o;    double r;    Circle(){}    Circle(Point a,double b):o(a),r(b){}    double area(){return sqr(r)*PI;}    double len(double ang){return r*ang;}};struct Tri{    Point p[3];};typedef Point Vector;double cross(Vector a,Vector b){    return a.x*b.y-a.y*b.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 dist(Point a,Point b){    return length(a-b);}double v_angle(Vector a,Vector b){    return acos(dot(a,b)/length(a)/length(b));}int ConvexHull(Point *p, int n, Point *ch){    sort(p, p+n);    n = unique(p, p+n) - p;    int m = 0;    for(int i = 0; i < n; i++)    {        while(m > 1 && dcmp(cross(ch[m-1]-ch[m-2], p[i]-ch[m-2])) <= 0) m--;        ch[m++] = p[i];    }    int k = m;    for(int i = n-2; i >= 0; i--)    {        while(m > k && dcmp(cross(ch[m-1]-ch[m-2], p[i]-ch[m-2])) <= 0) m--;        ch[m++] = p[i];    }    if(n > 1) m--;    return m;}Vector rotate(Vector a,double rad){    Vector c;    c.x = a.x*cos(rad)-a.y*sin(rad);    c.y = a.x*sin(rad)+a.y*cos(rad);    return c;}void get_ocmt(Circle c1,Circle c2,Point &s1, Point &e1,Point &s2,Point &e2){    double l = dist(c1.o,c2.o);    double d = fabs(c1.r-c2.r);    double theta = acos(d/l);    //if(dcmp(c1.r-c2.r)>0) swap(c1,c2);    Vector vec = c1.o-c2.o;    vec = vec.trunc(c1.r);    s1 = c1.o+rotate(vec,theta);    s2 = c1.o+vec.rotate(-theta);    vec = vec.trunc(c2.r);    e1 = c2.o+vec.rotate(theta);    e2 = c2.o+vec.rotate(-theta);}void get_pc(Circle c, Point p,Point &s1,Point &s2){    Vector u = p-c.o;    double dist = length(u);    Point v = c.o+u/dist*c.r;    double ang = PI/2-asin(c.r/dist);    s1 = rotate(v-c.o,-ang)+c.o;    s2 = rotate(v-c.o,ang)+c.o;}Point p[55*55*55],ch[55*55*55];int main(){    //freopen("/home/qitaishui/code/in.txt","r",stdin);    int n,m,pn,chn;    Circle c[MAXN];    Tri tri[MAXN];    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i = 0; i < n;i++)        {            c[i].o.input();            scanf("%lf",&c[i].r);        }        if(n==1&&m==0)        {            printf("%.6f\n",c[0].len(2*PI));            continue;        }        for(int i = 0; i < m; i++)            for(int j = 0; j < 3; j++)                tri[i].p[j].input();        pn = 0;        for(int i = 0; i < n; i++)            for(int j = i+1; j < n; j++)            {                if(dcmp(c[i].r-c[j].r)>0)                    get_ocmt(c[j],c[i],p[pn+1],p[pn],p[pn+3],p[pn+2]);                else                    get_ocmt(c[i],c[j],p[pn],p[pn+1],p[pn+2],p[pn+3]);                p[pn].tp = 0,p[pn].id = i;                p[pn+1].tp = 0,p[pn+1].id = j;                p[pn+2].tp = 0,p[pn+2].id = i;                p[pn+3].tp = 0,p[pn+3].id = j;                pn+=4;            }        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                for(int k = 0; k <3; k++)                {                    get_pc(c[i],tri[j].p[k],p[pn],p[pn+1]);                    p[pn].tp = 0,p[pn].id = i;                    p[pn+1].tp = 0,p[pn+1].id = i;                    pn+=2;                }        for(int j = 0; j < m; j++)            for(int k = 0; k <3; k++)            {                p[pn] = tri[j].p[k];                p[pn].tp = 1, p[pn].id = j;                pn++;            }        chn = ConvexHull(p,pn,ch);        //cout<<chn<<endl;        int top=0;        bool flag = 0;        double ans = 0,cir=0;        for(int i = 0; i <chn; i++)        {            if(ch[i].tp==0&&ch[(i+1)%chn].tp==0&&ch[i].id==ch[(i+1)%chn].id)            {                int tmp=ch[i].id;                double ang;                ang = v_angle(ch[i]-c[tmp].o,ch[(i+1)%chn]-c[tmp].o);                if(dcmp(cross(ch[i]-c[tmp].o,ch[(i+1)%chn]-c[tmp].o))<0)                    ang = 2*PI-ang;                ans=ans+c[tmp].len(ang);            }            else                ans+=length(ch[i]-ch[(i+1)%chn]);        }        printf("%.6f\n",ans);    }    return 0;}