入门题，给了A，B，C三点的坐标，通过参数方程可以求得，在BC上点的坐标 p = B + ( C - B ) t , 因为D是三等分点，所以t = 1 / 3. 可以求得D = 2 / 3 * B + 1 / 3 * C，同理可以求得E，F。
这样就可以求得点P， Q， R三点的坐标。例如求P点坐标方法如下
Point GetLineIntersection(Point p, Vector v, Point Q, Vector w) { //求两条直线的交点Vector u = p - Q;double t = Cross(w, u) / Cross(v, w);return p + v * t;
}
Point getP(Point A, Point B, Point C) {Point D = B * 2 / 3 + C / 3;Vector AD = D - A;Point E = C * 2 / 3 + A / 3;Vector BE = E - B;return GetLineIntersection(A, AD, B, BE);
} 求三角形面积的方法就好多了，我提供两种方式。
一 用的是求出QR的距离，在求出P点到直线QR的距离（注意是直线），这种方法。
二 调用Aera2这个函数。
/*************************************************************************> File Name: UVa11437.cpp> Author: AcToy> Mail: ycsgldy@163.com > Created Time: 2013年07月18日 星期四 17时48分55秒************************************************************************/#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include using namespace std;
typedef unsigned int u32;
typedef long long i64;
typedef unsigned long long u64;
typedef vector IV;
typedef vector BV;
typedef pair II;
typedef vector IIV;
#define For(t,v,c) for(t::const_iterator v=c.begin(); v!=c.end(); ++v)
const int INF = 0x7FFFFFFF;
const double eps = 1E-10;
const double PI = acos(-1);
struct Point {double x, y;Point(double x = 0, double y = 0) : x(x), y(y) { }
};
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }
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 Area2(Point A, Point B, Point C) { return Cross(B - A, C - A); }
int dcmp(double x) {if(fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point &b) {return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
Point GetLineIntersection(Point p, Vector v, Point Q, Vector w) {Vector u = p - Q;double t = Cross(w, u) / Cross(v, w);return p + v * t;
}
double DistanceToLine(Point P, Point A, Point B) {Vector v1 = B - A, v2 = P - A;return fabs(Cross(v1, v2)) / Length(v1);
}
Point read_point() {double x, y;scanf("%lf%lf", &x, &y);return Point(x, y);
}
Point getP(Point A, Point B, Point C) {Point D = B * 2 / 3 + C / 3;Vector AD = D - A;Point E = C * 2 / 3 + A / 3;Vector BE = E - B;return GetLineIntersection(A, AD, B, BE);
}
Point getR(Point A, Point B, Point C) {Point D = B * 2 / 3 + C / 3;Vector AD = D - A;Point F = A * 2 / 3 + B / 3;Vector CF = F - C;return GetLineIntersection(A, AD, C, CF);
}
Point getQ(Point A, Point B, Point C) {Point F = A * 2 / 3 + B / 3;Vector CF = F - C;Point E = C * 2 / 3 + A / 3;Vector BE = E - B;return GetLineIntersection(B, BE, C, CF);
}
int main() {int Case;Point A, B, C, D, E, F, P, Q, R;scanf("%d", &Case);while(Case--) {A = read_point(), B = read_point(), C = read_point();P = getP(A, B, C), R = getR(A, B, C), Q = getQ(A, B, C);double cnt = Length(R - Q);printf("%.0lf\n", cnt * DistanceToLine(P, R, Q) / 2);//printf("%.0lf\n", Area2(P, Q, R) / 2);}return 0;
}
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