HDU 5295 Unstable

problem

题意

• 题意非常简单，给定AB,BC,CD,DA,EF五条线段的长度，确定这五个点的坐标。其中E是AB的中点，F是CD的中点。有SPJ，所以只要满足这五条线段长度的点就可以了。

思路

• 非常纯净的一道几何题。。我们假设我们已经得到了这个四边形。对这个四边形做先做以下的操作：连接AF，倍长至A’。连接A’B,A’C,过A’做BC平行线，过B做A’C平行线,交于G。连接GD。如下图。
  
• 从这个图形我们可以得到一些性质。观察三角形BCA‘,我们可以知道，这个三角形的三条边分别为BC,AD,2*EF，也就是说这个三角形是确定的。我们从这里入手，先确定BC两点的坐标（任意即可），然后以B为圆心做一个半径为2EF的圆，以C为圆心做一个长度为DA的圆。这样我们就确定了A‘。再根据A‘来确定G点的坐标。然后以G为圆心，做一个半径为AB的圆，以C为圆心，确定长度为CD的圆。这两圆的交点为D。然后A也可以求出来了。

AC代码

HDU5295

/*************************************************************************
  > File Name: main.cpp
  > Author: znl1087
  > Mail: loveCJNforever@gmail.com 
  > Created Time: 日  8/ 9 13:33:25 2015
 ************************************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1.5e-6;
struct Point{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
};
typedef Point Vec;
int dcmp(double x){
    if(fabs(x) < EPS) return 0; else return x < 0 ? -1: 1;
}
Vec operator + (Vec a,Vec b){return Vec(a.x + b.x, a.y + b.y);}
Vec operator - (Vec a,Vec b){return Vec(a.x - b.x, a.y - b.y);}
Vec operator * (Vec a,double p){return Vec(a.x * p,a.y * p);}
Vec operator / (Vec a,double p){return Vec(a.x / p,a.y / p);}
bool operator == (const Point& a,const Point& b){
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y)==0;
}
double angle(Vec v){return atan2(v.y,v.x);}

double Dot(Vec a,Vec b){return a.x * b.x + a.y * b.y;}
double Length(Vec a){return sqrt(Dot(a,a));}
double Angle(Vec a,Vec b){return acos(Dot(a,b)/Length(a)/Length(b));}
double Cross(Vec a,Vec b){return a.x * b.y - a.y * b.x;}

struct Circle{
    Point c;
    double r;
    Circle(){}
    Circle(Point c,double r):c(c),r(r){}
    Point point(double a){
        return Point(c.x + cos(a) * r , c.y + sin(a) * r);
    }
};

int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol){
    double d = Length(C1.c - C2.c);
    if( dcmp(C1.r - C2.r)>0)
        swap(C1,C2);
    if(dcmp(d) == 0){
        if(dcmp(C1.r - C2.r)==0){
            sol.push_back(C1.c + Vec(0,C1.r));
            sol.push_back(C1.c - Vec(0,C1.r));
            return 2;
        }
        return 0;
    }
    if(dcmp(C1.r + C2.r - d) < 0) return 0;
    if(dcmp(fabs(C1.r - C2.r)-d) > 0)return 0;
    if(dcmp(C1.r + C2.r - d) == 0 || dcmp(fabs(C1.r - C2.r)-d) == 0){
        Vec p = C1.c-C2.c;
        sol.push_back(C2.c + p / Length(p) * C2.r);
        return 1;
    }
    double a = angle(C2.c - C1.c);
    double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
    Point p1 = C1.point(a-da),p2 = C1.point(a+da);
    sol.push_back(p1);
    if(p1 == p2)return 1;
    sol.push_back(p2);
    return 2;
}
double AB,BC,CD,DA,EF;
int main(){
    int T;
    cin>>T;
    for(int cas = 1;cas <= T;cas++){
        scanf("%lf%lf%lf%lf%lf",&AB,&BC,&CD,&DA,&EF);
        Circle B = Circle(Point(0,0),2*EF),C = Circle(Point(0,BC),DA),G,D;
        Point A,A2;
        vector<Point> ans;
        int ok = getCircleCircleIntersection(B, C,ans);
        for(int ook = 0;ook<ok;ook++){
            A2 = ans[ook];
            G = Circle(A2 + (B.c - C.c),AB);
            C.r = CD;
            ans.clear();
            int flag = getCircleCircleIntersection(C, G, ans);
            if(flag == 0)continue;
            D = Circle(ans[0],DA);
            //printf("GD:%lf\n",Length(G.c-D.c));
            A = B.c + (D.c-G.c);
            printf("Case #%d:\n",cas);
            printf("%.6f %.6f\n",A.x,A.y);
            printf("%.6f %.6f\n",B.c.x,B.c.y);
            printf("%.6f %.6f\n",C.c.x,C.c.y);
            printf("%.6f %.6f\n",D.c.x,D.c.y);
            Point E = (A + B.c)/2.0,F = (C.c + D.c)/2.0;
            /*printf("AB:%lf BC:%lf CD:%lf DA:%lf EF:%lf\n",
                    Length(A-B.c),Length(B.c-C.c),Length(C.c-D.c),
                    Length(D.c-A),Length(E-F));*/
            break;
        }
    }
    return 0;
}