Question

给出2个三角形边长。确定第二个三角形是否适合第一个三角形？

有关详细信息，请阅读以下完整的问题陈述：

http://acm.timus.ru/problem.aspx?space=1&num=1566&locale=en

下面的实现尝试了对齐三角形底边的所有（3！）^ 2种可能的组合。然后它会尝试在第一个三角形内移动第二个三角形，同时检查第二个三角形的底边是否超过第一个三角形的底边。

但我一直得到错误答案（WA）＃16。

enter image description here

我给出的案例是第二张图片。很明显，如果旋转PQR以对齐长度为2.77和3.0的边，则第三个顶点将不在三角形ABC内。长度为4.2的边只能沿len 5的边对齐。因此，只有在第二张图像中显示的配置才能满足此情况。

你能帮我找到这个bug，建议一些我的算法崩溃的测试用例。也欢迎替代算法。

#include <cmath>
#include <iostream>
using namespace std;

const double PI = atan(1.0)* 4;

// Traingle ABC (Envelope)
double xa, ya, xb, yb, xc, yc;

// Traingle PQR (Postcard)
double xp, yp, xq, yq, xr, yr;

// Angle between sides AB and AC
double theta;

double signWrtLine(double x1, double y1, double x2, double y2, double x, double y)
{
    double A = y2 - y1;
    double B = x1 - x2;
    double C = -(A * x1 + B * y1);

    return (A * x + B * y + C);
}

bool fit()
{ 
    if ((xr > xc) || (yq > yb)) return false;

    if (signWrtLine(xa, ya, xb, yb, xq, yq) < 0) {
        double d = (yq / tan(theta)) - xq;
        return (xr + d <= xc);
    }

    return (signWrtLine(xa, ya, xb, yb, xq, yq) >= 0 && 
            signWrtLine(xb, yb, xc, yc, xq, yq) >= 0 && 
            signWrtLine(xc, yc, xa, ya, xq, yq) >= 0);
}

bool fit(double a[], double b[])
{
    // generate the 3! permutations of the envelope
    // loops i,k
    for (int i = 0; i < 3; i++) {

        double angle;
        double u = a[i], v = a[(i + 1) % 3], w = a[(i + 2) % 3];

        for (int k = 0; k < 2; k++) {
            switch (k) {
            case 0:
                xa = 0, ya = 0;
                angle = theta = acos((u * u + v * v - w * w) / (2 * u * v));
                xb = v * cos(angle), yb = v * sin(angle);
                xc = u, yc = 0;     
                break;
            case 1:
                // reflect envelope
                swap(v, w);
                angle = theta = acos((u * u + v * v - w * w) / (2 * u * v));
                xb = v * cos(angle), yb = v * sin(angle);       
                break;
            }

            // generate the 3! permutations of the postcard
            // loops j,k
            for (int j = 0; j < 3; j++) {

                double angle;
                double u = b[j], v = b[(j + 1) % 3], w = b[(j + 2) % 3];

                for (int k = 0; k < 2; k++) {
                    switch (k) {
                    case 0:
                        xp = 0, yp = 0;
                        angle = acos((u * u + v * v - w * w) / (2 * u * v));
                        xq = v * cos(angle), yq = v * sin(angle);
                        xr = u, yr = 0;
                        break;
                    case 1:
                        // reflect postcard
                        swap(v, w);
                        angle = acos((u * u + v * v - w * w) / (2 * u * v));
                        xq = v * cos(angle), yq = v * sin(angle);
                        break;
                    }

                    if (fit()) return true;
                }
            }
        }
    }
    return false;
}


int main()
{
    double a[3], b[3];

    for (int i = 0; i < 3; i++) cin >> a[i];
    for (int i = 0; i < 3; i++) cin >> b[i];

    if(fit(a, b)) cout << "YES" << endl;
    else cout << "NO" << endl;

    return 0;
}

Answer 1

重心坐标！详细说明：

让“包络”三角形具有顶点A，B，C;在不失一般性的情况下，您可以将顶点A放置在原点，并将AB侧与+ x轴对齐。使用包络三角形的边长来找到顶点A处的角度，即边AB和AC之间的角度。使用此角度，您可以定义新的坐标系（u，v）;在这个坐标系中，顶点坐标是A =（0,0），B =（1,0）和C =（0,1）。

现在，取另一个带顶点A'，B'，C'的三角形，首先找到3个顶点的XY坐标：（A'B'，B'C'，A'C' ）与+ x坐标轴对齐。对于每个这样的对齐，将其他两个顶点转换为由包络三角形确定的UV坐标系。如果发生两个其他顶点具有（u，v）坐标，其中0 <= u，v <= 1且u + v <= 1，则三角形适合包络三角形。

两面之间的角度可以通过平面三角形的正弦定理获得;虽然你必须要小心，如果顶点的角度是钝角（> PI / 2），因为正弦函数在区间[0，PI]上围绕PI / 2对称。要检查角度是否为钝角，您还需要使用余弦定理，尽管您不需要计算余弦本身：if | AB | ^ 2 + | AC | ^ 2＆gt; | BC | ^ 2，A处的角度是钝的。

我认为总结一下。

Answer 2

//23/08/11 13:56
//determine if a triangle will fit inside a second triangle
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const double pi= 3.1414926;
const double tri=180;//deg in triangle
double find_B_ang(double a,double b,double c);
double find_C_ang(double a,double b,double c);
double movetri_r , tc_ghosthor_B;
int main()
{double a=0.0,b=0.0,c=0.0,x=0.0,y=0.0,z=0.0;
double A=0.0,B=0.0,C=0.0,A1=0.0,B1=0.0,C1=0.0;// L&R base angles
double te_vert_B=0.0,te_hor_B=0.0,te_hor_C=0.0;
double tc_vert_B=0.0,tc_hor_B=0.0,tc_hor_C=0.0;
//points B and B1 are considered co-incedent
cout<<"\n\ndetermine if a triangular card will fit inside\n"
    <<"a triangular envelope\n";
//envelope dimensions    
cout<<"\nenter lengths of the sides of envelope (space between)\n";
cout<<"ensure longest of them is less than sum of other two\n";
do
{
   cin>>a>>b>>c;//the e sides
   if(c>a)swap(a,c);//sort sides in decending order
   if(b>a)swap(a,b);
   if(c>b)swap(b,c);
   if(a >(b+c))
   cout<<"WRONG...b+c must be greater than a";
}while(a >(b+c));


cout<<"\nthe sides of the envelope are "<<a<<','<<b<<','<<c<<endl;
B=find_B_ang(a,b,c);
C=find_C_ang(a,b,c);
te_vert_B=c*sin(B*pi/tri);//apex to base vertical line
te_hor_B=te_vert_B/tan(B*pi/tri);//B to vertical line
te_hor_C=a-te_hor_B;//C to vertical line

cout<<"-------------------------------------------\n";
//card dimensions
do
{
cout<<"\nenter lengths of sides of card (space between) \n"; 
cout<<"ensure longest of them is less than sum of other two\n";
do
{
   cin>>x>>y>>z;//the c sides
   if(z>x)swap(z,x);//sort sides in decending order
   if(y>x)swap(y,x);
   if(z>y)swap(y,z);
   if(x>(y+z))
   cout<<"WRONG...y+z must be greater than x\n";
}while(x>(y+z));

cout<<"\nthe sides of card are "<<x<<','<<y<<','<<z<<endl;//x is base
B1=find_B_ang(x,y,z);
C1=find_C_ang(x,y,z);
tc_vert_B=z*sin(B1*pi/tri);//apex to base vertical line
tc_hor_B=tc_vert_B/tan(B1*pi/tri);//B to vertical line
tc_hor_C=x-tc_hor_B;//C to vertical line
tc_ghosthor_B=tc_vert_B/tan(B*pi/tri);
movetri_r= abs(tc_ghosthor_B-tc_hor_B);    
cout<<"------------------------------------------------\n";
//determine and advise if card fits within envelope    
if(B1<B && tc_vert_B <(tc_hor_C + a-x)*tan(C*pi/tri))cout<<"\ntrue";
else if(B1<B && tc_hor_B< te_hor_B && tc_vert_B<te_vert_B)cout<<"true";
else if(B1>B && movetri_r<a-x && tc_vert_B<te_vert_B)cout<<"true";
else cout<<"\nfalse";
} while(x>0);
cout<<"\npress any key...";
cin.ignore();
cin.get();
return 0;
}
double find_B_ang(double a,double b,double c)
{
 double X=0.0;
 X=((a*a)+(c*c)-(b*b));
 X/=2*a*c;
 X=acos(X);
 X*=tri/pi;
 return X; //degrees
}
double find_C_ang(double a,double b,double c)
{
 double X=0.0;
 X=((a*a)+(b*b)-(c*c));
 X/=2*a*b;       
 X=acos(X);
 X*=tri/pi;
 return X;//degrees
}

Answer 3

可以从这里尝试 - http://www.springerlink.com/content/t10266u5832477w7/。到目前为止，问题似乎尚未得到解决，所以最好选择一些启发式方法来获得简单的案例（例如检查刻录/限定圆圈，对齐边界等）并希望获得最佳效果。

Answer 4

比较双打时使用epsilon（1e-10）！

三角形适合另一个三角形

4 个答案: