[编程题] 寻找三角形

时间限制：1秒空间限制：32768K

三维空间中有N个点，每个点可能是三种颜色的其中之一，三种颜色分别是红绿蓝，分别用’R’, ‘G’, ‘B’表示。
现在要找出三个点，并组成一个三角形，使得这个三角形的面积最大。
但是三角形必须满足：三个点的颜色要么全部相同，要么全部不同。
输入描述:
首先输入一个正整数N三维坐标系内的点的个数.(N <= 50)

接下来N行，每一行输入 c x y z，c为’R’, ‘G’, ‘B’ 的其中一个。x，y，z是该点的坐标。(坐标均是0到999之间的整数)

输出描述:
输出一个数表示最大的三角形面积，保留5位小数。

输入例子:

5R 0 0 0R 0 4 0R 0 0 3G 92 14 7G 12 16 8

输出例子:

6.00000

代码：

#include<iostream>#include <math.h>#include <iomanip>#define MAX_NUM 50using namespace std;struct Triangle{    int x = 0;    int y = 0;    int z = 0;    char c = 'N';};bool isTriangle(Triangle tri1, Triangle tri2, Triangle tri3){    //颜色判断    if ((tri1.c == tri2.c && tri2.c == tri3.c) || (tri1.c != tri2.c && tri1.c != tri3.c && tri2.c != tri3.c)){        //判断是否共线        Triangle t1, t2;        t1.x = tri1.x - tri2.x;        t1.y = tri1.y - tri2.y;        t1.z = tri1.z - tri2.z;        t2.x = tri1.x - tri3.x;        t2.y = tri1.y - tri3.y;        t2.z = tri1.z - tri3.z;        double temp1 = t1.x * t2.x + t1.y + t2.y + t1.z* t2.z;        double temp2 = sqrt((pow(t1.x, 2) + pow(t1.y, 2) + pow(t1.z, 2))*(pow(t2.x, 2) + pow(t2.y, 2) + pow(t2.z, 2)));        if (temp1 == temp2 || temp1 + temp2 == 0){            return false;        }        else{            return true;        }    }    else{        return false;    }}double getLine(Triangle tri1, Triangle tri2){    return sqrt(pow((tri1.x - tri2.x), 2) + pow((tri1.y - tri2.y), 2) + pow((tri1.z - tri2.z), 2));}double getArea(Triangle tri1, Triangle tri2, Triangle tri3){    double a, b, c;    a = getLine(tri1, tri2);    b = getLine(tri1, tri3);    c = getLine(tri2, tri3);    double p = (a + b + c) / 2;    double temp = p * (p - a)*(p - b)*(p - c);    double S = sqrt(temp);    return S;}int main(){    int N;    Triangle tri[MAX_NUM];    cin >> N;    for (int i = 0; i < N; i++){        cin >> tri[i].c;        cin >> tri[i].x;        cin >> tri[i].y;        cin >> tri[i].z;    }    double max = 0;    for (int i = 0; i < N; i++){        for (int j = i + 1; j < N; j++){            for (int k = j + 1; k < N; k++){                if (isTriangle(tri[i], tri[j], tri[k])){                    double temp = getArea(tri[i], tri[j], tri[k]);                    if (temp>max)                        max = temp;                }            }        }    }    cout << setiosflags(ios::fixed) << setprecision(5) << max << endl;    return 0;}

*C++输出固定小数位数：cout << setiosflags(ios::fixed) << setprecision(5) << max << endl;
精确度的话去掉fixed即可。