EOJ 1594. Triangle

/* 一道艰难的实训题 */

单点时限: 2.0 sec

内存限制: 256 MB

A triangle is a basic shape of planar geometry. It consists of three straight lines and three angles in between. Sides are usually labeled a,b,c, with their opposite angles labeled α,β,γ respectively.

A look into a book about geometry shows that the following formulas exists:

The values of a,b,c,α,β,γ forms a set of six parameters that fully define a triangle. If a large enough set of parameters are given, the missing ones can be calculated using the formulas above.

You are to write a program that calculates the missing parameters for a given subset of the six parameters of a triangle. For some sets of parameters, it is not possible to calculate the triangle because either too few is known about the triangle or the parameters would lead to an invalid triangle. The sides of a valid triangle are greater than 0 and the angles are greater than 0 and less than π. Your program should detect this case and output: Invalid input. The same phrase should be output if more than the minimal set needed to compute the triangle is given but the parameters conflict with each other, e.g. all three angles are given but their sum is greater than π.

Other sets of parameters can lead to more than one but still a finite number of valid solutions for the triangle. In such a case, your program should output: “More than one solution.”

In all other cases, your program should compute the missing parameters and output all six parameters.

输入格式
The first line of the input file contains a number indicating the number of parameter sets to follow. Each following line consists of six numbers, separated by a single blank character. The numbers are the values for the parameters a,α,b,β,c,γ respectively. A value of −1 indicates that the corresponding parameter is undefined and has to be calculated. All floating-point numbers include at least eight significant digits.

输出格式
Your program should output a line for each set of parameters found in the input file. If a unique solution for a valid triangle can be found for the given parameters, your program should output the six parameters a,α,b,β,c,γ separated by a blank character. Otherwise the line should contain the phrase More than one solution. or Invalid input. as explained above.

The numbers in the output file should include at least six significant digits. Your calculations should be precise enough to get the six most significant digits correct (i.e. a relative error of 0.000001 is allowed).

样例

input
4 47.9337906847 0.6543010109 78.44555175791.4813893731 66.5243757656 1.0059022695 62.72048064 2.26853639 -1.00000000 0.56794657 -1.00000000 -1.00000000 15.69326944 0.24714213 -1.00000000 1.80433105 66.04067877 -1.00000000 72.83685175 1.04409241 -1.00000000 -1.00000000 -1.00000000 -1.00000000
output
47.933791 0.654301 78.445552 1.481389 66.524376 1.005902 62.720481 2.268536 44.026687 0.567947 24.587225 0.305110 Invalid input. Invalid input.

解题思路
比较棘手的一道题，题意就是通过给出的条件解出−1-1−1所代表的元素，初中解三角形的功底要扎实，对各种情况分类讨论。

无效输入的情况：
① 题目所给的数据少于三个，无法得到一个解（计算前判断）
② 要素无法构成三角形(judge函数，计算后判断)：
1. 仍有未知项（即-1）
2. 三个角之和不等于π\piπ
3. 两边之和小于第三边
4. 不满足正弦定理
求解三角形：
判断解的个数：
已知两边和其中一边的对角不能唯一确定三角形，可能出现两解，一解或无解的情况。在本题中使用正弦定理来判断。
若已知 a、b、αa、b、\alphaa、b、α，由正弦定理可以得到sinβ=bsinαasin\beta=\frac{bsin\alpha}{a}sinβ=absinα，若sinβ>1sin\beta>1sinβ>1，无解；若sinβ=1sin\beta=1sinβ=1，一解；若sinβ<1sin\beta<1sinβ<1两解。在代码中则通过比较 aaa 与 bsinαbsin\alphabsinα 的大小来判断。

代码思路
写代码的时候没有怎么打注释，所以解释一下。
由于题目有精度要求，所以宏定义一个小数mask。
在输入时指定ccc和eee来分别储存已知的角和边的个数。
首先定义三个函数

已知三边时调用getagetageta函数求角
已知两边及其夹角时调用getegetegete函数求对边
c=2c=2c=2 时调用 c2cc2cc2c 函数求第三个角
e=3e=3e=3 时调用 e2ce2ce2c 求所有的角

之后进入主函数

读入数据并记录已知元素个数；
判断元素数量是否满足最低限度；
仅根据已知边或角的数量来处理：已知两角求第三个角或已知三边求所有角；
边角情况混合处理：
（1）已知一边及其对角(a,α)(a,\alpha)(a,α)，此时可以解出来的情况有(−1,β)(-1,\beta)(−1,β)或(b,−1)(b,-1)(b,−1)；
①(−1,β)(-1,\beta)(−1,β)，用正弦定理求出bbb;
②(b,−1)(b,-1)(b,−1)，此时需要根据 sinβsin\betasinβ 来判断解的情况；
（2）已知两边和未知边的对角，直接调用 getegetegete 函数解出第三边；
至此我们好像已经遍历了所有情况，只需要循环至所有元素都被解出来；
调用 judgejudgejudge 函数判断元素是否可以组成三角形，判断是否有多解；
输出。

ac代码

#include <bits/stdc++.h>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
const long double pi=4*atan(1);
const long double mask=1e-6;
long double a[6];long double geta(long double a,long double b,long double c)
{return acos((a*a+b*b-c*c)/(2*a*b));
}long double gete(long double x,long double y,long double c)
{return sqrt(x*x+y*y-2*x*y*cos(c));
}void c2c()
{long double t=0;for(int i=1;i<6;i+=2){if(a[i]<0)continue;t+=a[i];}for(int i=1;i<6;i+=2)if(a[i]<0)a[i]=pi-t;
}void e2c()
{for(int i=1;i<6;i+=2)if(a[i]<0)a[i]=geta(a[(i+1)%6],a[(i+3)%6],a[i-1]);
}bool judge()
{for(int i=0;i<6;i++)if(a[i]<0)return false;if(fabs(a[1]+a[3]+a[5]-pi)>mask)return false;if(a[0]+a[2]<=a[4]||a[2]+a[4]<=a[0]||a[0]+a[4]<=a[2])return false;for (int i = 0; i < 4; i += 2) if (fabs(a[i]*sin(a[i+3])-sin(a[i+1])*a[i+2])>mask)return false;return true;
}int main(int argc, char** argv)
{int T;cin >> T;while(T--){int n=0,c=0,e=0;bool flag=false;for(int i=0;i<6;i++){cin >> a[i];if(a[i]>0){n++;if(i&1) c++;else e++;}}if(n<3){cout << "Invalid input." << endl;continue;}while(n<6){int s=n;if(c==2){c2c();n++;c=3;}if(e==3){e2c();n+=3-c;c=3;}for(int i=0;i<6;i+=2){if(a[i]>0&&a[i+1]>0){for(int j=0;j<6;j+=2){if((a[j]>0&&a[j+1]>0)||(a[j]<0&&a[j+1]<0))continue;if(a[j]<0){a[j]=a[i]/sin(a[i+1])*sin(a[j+1]);n++;e++;}else{if(a[i+1]<pi/2&&a[i]<a[j]&&a[i]>a[j]*sin(a[i+1]))flag=true;double temp=sin(a[i+1])*a[j]/a[i];if(temp>1||temp<0)continue;a[j+1]=asin(sin(a[i+1])*a[j]/a[i]);n++;c++;}}}}if(e==2){for(int i=0;i<6;i+=2){if(a[i]<0&&a[i+1]>0){a[i]=gete(a[(i+2)%6],a[(i+4)%6],a[i + 1]);n++;e++;}}}if(s==n)break;}if(judge()){if(flag)cout << "More than one solution." << endl;else{for(int i=0;i<5;i++)cout << fixed << setprecision(6) << a[i] << " ";cout << fixed << setprecision(6) << a[5] << endl;}}elsecout << "Invalid input." << endl;}return 0;
}

补充
有位带哥提醒我 π\piπ 可以用M_PI表示，更方便些

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