hdu 5295 Unstable（计算几何）

题目链接：hdu 5295 Unstable

先确定B，C点，再找到A‘点A’C = AD，A'B = 2EF，然后确定G点向量A'G = CB（注意是向量）。然后根据DG = AB确定D点。三角形DAF和FBA‘为全等三角形，所以A点也可以得到。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const double eps = 1e-9;

inline int sgn(double x) {
return x < 0 ? -1 : 1;
}

struct Point {
double x, y;
Point (double x = 0, double y = 0) {
set(x, y);
}
void set(double x, double y) {
this->x = x;
this->y = y;
}
void put() {
printf("%lf %lf\n", x, y);
}
Point operator - (const Point& a) {
Point ret;
ret.set(x - a.x, y - a.y);
return ret;
}
Point operator + (const Point& a) {
Point ret;
ret.set(x + a.x, y + a.y);
return ret;
}
bool operator == (const Point& a) {
return fabs(x - a.x) < eps && fabs(y - a.y) < eps;
}
};

struct Circle {
Point o;
double r;
void set(Point o, double r) {
this->o = o;
this->r = r;
}
bool operator == (const Circle& a) {
return o == a.o && fabs(r - a.r) < eps;
}
};

Circle P, Q;
Point A, A2, B, C, D, G, T;
double AB, BC, CD, DA, EF;

void getPoint(Circle p, Circle q, Point& u, Point& v) {
if (p == q) {
u = p.o + Point(0, p.r);
v = p.o - Point(0, p.r);
return;
}

double a = q.o.x - p.o.x, b = q.o.y - p.o.y;
double c = (pow(p.r, 2) - pow(q.r, 2) + a * a + b * b) / 2;

if (fabs(a) > eps && fabs(b) > eps) {
double delta = b * b * c * c - (a * a + b * b) * (c * c - p.r * p.r * a * a);
if (sgn(delta) <= 0)
delta = 0;
u.y = (b * c + sqrt(delta)) / (a * a + b * b);
v.y = (b * c - sqrt(delta)) / (a * a + b * b);
u.x = (c - b * u.y) / a;
v.x = (c - b * u.y) / a;
} else if (fabs(a) > eps && fabs(b) < eps) {
u.x = v.x = c / a;
double tmp = p.r * p.r - u.x * u.x;
if (sgn(tmp) <= 0)
tmp = 0;
u.y = sqrt(tmp);
v.y = -u.y;
} else if (fabs(a) < eps && fabs(b) > eps) {
u.y = v.y = c / b;
double tmp = p.r * p.r - u.y * u.y;
if (sgn(tmp) <= 0)
tmp = 0;
u.x = sqrt(tmp);
v.x = -u.x;
}
u = u + p.o;
v = v + p.o;

}

void solve () {
B.set(0, 0);
C.set(BC, 0);
P.set(B, 2 * EF);
Q.set(C, DA);
getPoint(P, Q, A2, T);

G = A2 + (B - C);

P.set(G, AB);
Q.set(C, CD);

getPoint(P, Q, D, T);
A = D + (C - A2);
}

int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
scanf("%lf%lf%lf%lf%lf", &AB, &BC, &CD, &DA, &EF);
printf("Case #%d:\n", kcas);
solve();
A.put();
B.put();
C.put();
D.put();
}
return 0;
}