HDU 4720 Naive and Silly Muggles(几何)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4720

用几何模板，求外接圆，再判断点在不在圆内

#include <stdio.h>
#include <string.h>
#include <math.h>
const double esp = 1e-9;

//点
struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) {
	this->x = x;
	this->y = y;
    }
    void read() {
	scanf("%lf%lf", &x, &y);
    }
};

//线段
struct Line {
    Point p[2];
    double A, B, C;
    Line() {}
    Line(Point a, Point b) {
	this->p[0] = a;
	this->p[1] = b;
	A = p[0].y - p[1].y;
	B = p[1].x - p[0].x;
	C = -(A * p[0].x + B * p[0].y);
    }
};

//圆
struct Circle {
    Point center;
    double r;
    Circle() {}
    Circle(Point center, double r) {
	this->center = center;
	this->r = r;
    }
};

//三角形
struct Tra {
    Point p[3];
    Tra() {}
    Tra(Point a, Point b, Point c) {
	this->p[0] = a;
	this->p[1] = b;
	this->p[2] = c;
    }
};

//点点距离
double dis(Point a, Point b) {
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}

//点线对称点
Point P_dc(Line l, Point p) {
    Point ans;
    ans.x = p.x - (2 * l.A * (l.A * p.x + l.B * p.y + l.C)) / (l.A * l.A + l.B * l.B);
    ans.y = p.y - (2 * l.B * (l.A * p.x + l.B * p.y + l.C)) / (l.A * l.A + l.B * l.B);
    return ans;
}

//三角形面积
double Tra_Area(Tra t) {
    return fabs((t.p[1].x - t.p[0].x) * (t.p[2].y - t.p[0].y) - (t.p[2].x - t.p[0].x) * (t.p[1].y - t.p[0].y)) / 2.0;
}

//求三角形外接圆
Circle Tra_Cir(Tra t) {
    Circle ans;
    double a = dis(t.p[0], t.p[1]);
    double b = dis(t.p[1], t.p[2]);
    double c = dis(t.p[2], t.p[0]);
    ans.r = (a * b * c) / (Tra_Area(t) * 4.0);
    double xA = t.p[0].x;
    double yA = t.p[0].y;
    double xB = t.p[1].x;
    double yB = t.p[1].y;
    double xC = t.p[2].x;
    double yC = t.p[2].y;
    double c1 = (xA*xA+yA*yA - xB*xB-yB*yB) / 2;
    double c2 = (xA*xA+yA*yA - xC*xC-yC*yC) / 2;
    ans.center.x = (c1*(yA - yC)-c2*(yA - yB)) / ((xA - xB)*(yA - yC)-(xA - xC)*(yA - yB));
    ans.center.y = (c1*(xA - xC)-c2*(xA - xB)) / ((yA - yB)*(xA - xC)-(yA - yC)*(xA - xB));
    return ans;
}

//判断点在圆内（含边界)
bool In_Cir(Circle c, Point p) {
    double d = dis(p, c.center);
    if (d - c.r > esp) return false;
    return true;
}

int t;
Point p[4];

bool judge(Circle pc) {
    if (!In_Cir(pc, p[3])) return true;
    Circle pc1 = Circle(P_dc(Line(p[0], p[1]), p[3]), pc.r);
    if (In_Cir(pc1, p[2]) && !In_Cir(pc1, p[3])) return true;
    pc1 = Circle(P_dc(Line(p[1], p[2]), p[3]), pc.r);
    if (In_Cir(pc1, p[0]) && !In_Cir(pc1, p[3])) return true;
    pc1 = Circle(P_dc(Line(p[0], p[2]), p[3]), pc.r);
    if (In_Cir(pc1, p[1]) && !In_Cir(pc1, p[3])) return true;
    return false;
}

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
	for (int i = 0; i < 4; i++)
	    p[i].read();
	Tra pt = Tra(p[0], p[1], p[2]);
	Circle pc = Tra_Cir(pt);
	printf("Case #%d: ", ++cas);
	if (judge(pc))
	    printf("Safe\n");
	else
	    printf("Danger\n");
    }
    return 0;
}

HDU 4720 Naive and Silly Muggles(几何),布布扣,bubuko.com

HDU 4720 Naive and Silly Muggles(几何)

原文：http://blog.csdn.net/accelerator_/article/details/25001285