模版
#define EPS (1e-10)
#define equals(a,b) (fabs((a) - (b)) < EPS)
// 点类
class Point {
public :
double x, y;
Point() {};
Point(double x, double y) :x(x), y(y) {}
Point operator + (Point p) { return Point(x + p.x, y + p.y); }
Point operator - (Point p) { return Point(x - p.x, y - p.y); }
Point operator * (double a) { return Point(x * a, y * a); }
Point operator / (double a) { return Point(x / a, y / a); }
bool operator < (const Point &p) const {
return x != p.x ? x < p.x : y < p.y;
}
bool operator == (const Point &p) const {
return fabs(x - p.x) < EPS && fabs(y - p.y) < EPS;
}
};
// 线段类
class Segment {
public:
Point p1, p2;
Segment() {};
Segment(Point p1, Point p2) :p1(p1), p2(p2) {};
};
// 圆类
class Circle {
public:
Point c;
double r;
Circle() {};
Circle(Point c, double r) :c(c), r(r) {}
};
// 定义向量
typedef Point Vector;
// 定义直线
typedef Segment Line;
// 定义多边形
typedef vector<Point> Polygon;
/***************************点、向量****************************/
double norm(Point p) { return p.x * p.x + p.y * p.y; }
double abs(Point p) { return sqrt(norm(p)); }
// 向量的内积
double dot(Point a, Point b) {
return a.x * b.x + a.y * b.y;
}
// 向量的外积
double cross(Point a, Point b) {
return a.x * b.y - a.y * b.x;
}
// 向量a,b是否正交 <==> 内积为0
bool isOrthogonal(Vector a, Vector b) {
return equals(dot(a, b), 0.0);
}
bool isOrthogonal(Point a1, Point a2, Point b1, Point b2) {
return equals(dot(a1 - a2, b1 - b2), 0.0);
}
// 向量a,b是否平行 <==> 外积为0
bool isParallel(Vector a, Vector b) {
return equals(cross(a, b), 0.0);
}
bool isParallel(Point a1, Point a2, Point b1, Point b2) {
return equals(cross(a1 - a2, b1 - b2), 0.0);
}
// 点p在线段s上的投影
Point project(Segment s, Point p) {
Vector base = s.p2 - s.p1;
double r = dot(p - s.p1, base) / norm(base);
return s.p1 + base * r ;
}
//以线段s为对称轴与点p成线对称的点
Point reflect(Segment s, Point p) {
return p + (project(s, p) - p) * 2.0;
}
// 点a到点b的距离
double getDistance(Point a, Point b) {
return abs(a - b);
}
// 线段l和点p的距离
double getDistanceLP(Line l, Point p) {
return abs( cross(l.p2 - l.p1, p - l.p1) / abs(l.p2 - l.p1) );
}
// 线段s与点p的距离
double getDistanceSP(Segment s, Point p) {
if (dot(s.p2 - s.p1, p - s.p1) < 0.0)
return abs(p - s.p1);
if (dot(s.p1 - s.p2, p - s.p2) < 0.0)
return abs(p - s.p2);
return getDistanceLP(s, p);
}
/*************************线段********************************/
// 线段s1,s2是否正交 <==> 内积为0
bool isOrthogonal(Segment s1, Segment s2) {
return equals(dot(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0);
}
// 线段s1,s2是否平行 <==> 外积为0
bool isParallel(Segment s1, Segment s2) {
return equals(cross(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0);
}
// 逆时针方向ccw(Counter-Clockwise)
static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;
int ccw(Point p0, Point p1, Point p2) {
Vector a = p1 - p0;
Vector b = p2 - p0;
if (cross(a, b) > EPS) return COUNTER_CLOCKWISE;
if (cross(a, b) < -EPS) return CLOCKWISE;
if (dot(a, b) < -EPS) return ONLINE_BACK;
if (norm(a) < norm(b)) return ONLINE_FRONT;
return ON_SEGMENT;
}
// 判断线段p1p2和线段p3p4是否相交
bool intersect(Point p1, Point p2, Point p3, Point p4) {
return (ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 &&
ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0);
}
//判断线段s1和s2是否相交
bool intersect(Segment s1, Segment s2) {
return intersect(s1.p1, s1.p2, s2.p1, s2.p2);
}
// 线段s1和线段s2的距离
double getDistance(Segment s1, Segment s2) {
// 相交
if (intersect(s1, s2))
return 0.0;
return min(min(getDistanceSP(s1, s2.p1), getDistanceSP(s1, s2.p2)),
min(getDistanceSP(s2, s1.p1), getDistanceSP(s2, s1.p2)));
}
// 线段s1与线段s2的交点
Point getCrossPoint(Segment s1, Segment s2) {
Vector base = s2.p2 - s2.p1;
double d1 = abs(cross(base, s1.p1 - s2.p1));
double d2 = abs(cross(base, s1.p2 - s2.p1));
double t = d1 / (d1 + d2);
return s1.p1 + (s1.p2 - s1.p1) * t;
}
/***************************圆****************************/
// 圆c和直线l的交点
pair<Point, Point> getCrossPoints(Circle c, Line l) {
Vector pr = project(l, c.c);
Vector e = (l.p2 - l.p1) / abs(l.p2 - l.p1);
double base = sqrt(c.r * c.r - norm(pr - c.c));
return make_pair(pr + e * base, pr - e * base);
}
// 圆c1和圆c2的交点
double arg(Vector p) { return atan2(p.y, p.x); }
Vector polar(double a, double r) { return Point(cos(r) * a, sin(r) * a); }
pair<Point, Point> getCrossPoints(Circle c1, Circle c2) {
double d = abs(c1.c - c2.c);
double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2 * c1.r * d));
double t = arg(c2.c - c1.c);
return make_pair(c1.c + polar(c1.r, t + a), c1.c + polar(c1.r, t - a));
}
/***************************多边形****************************/
// 点的内包
/*
IN 2
ON 1
OUT 0
*/
int contains(Polygon g, Point p) {
int n = g.size();
bool x = false;
for (int i = 0; i < n; i++) {
Point a = g[i] - p, b = g[(i + 1) % n] - p;
if (abs(cross(a, b)) < EPS && dot(a, b) < EPS) return 1;
if (a.y > b.y) swap(a, b);
if (a.y < EPS && EPS < b.y && cross(a, b) > EPS)
x = !x;
}
return (x ? 2 : 0);
}
int cmp(Point A, Point B) //竖直排序
{
return (A.y<B.y || (A.y == B.y&&A.x<B.x));
}
// 凸包
Polygon andrewScan(Polygon s) {
Polygon u, l;
int len = s.size();
if (len < 3) return s;
// 以x,y为基准升序排序
sort(s.begin(), s.end());
// 将x值最小的两个点添加到u
u.push_back(s[0]);
u.push_back(s[1]);
// 将x值最大的两个点添加到l
l.push_back(s[len - 1]);
l.push_back(s[len - 2]);
// 构建凸包上部
for (int i = 2; i < len; i++) {
for (int j = u.size(); j >= 2 && ccw(u[j - 2], u[j - 1], s[i]) >= 0; j--) {
u.pop_back();
}
u.push_back(s[i]);
}
// 构建凸包下部
for (int i = len - 3; i >= 0; i--) {
for (int j = l.size(); j >= 2 && ccw(l[j - 2], l[j - 1], s[i]) >= 0; j--) {
l.pop_back();
}
l.push_back(s[i]);
}
reverse(l.begin(), l.end());
for (int i = u.size() - 2; i >= 1; i--)
l.push_back(u[i]);
return l;
}
题目链接
求直线s1、s2的关系。若是平行则输出2,若相交则输出1,其他输出0
Point p1, p2, p3, p4;
int n;
scanf("%d", &n);
while (n--) {
scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &p1.x, &p1.y,
&p2.x, &p2.y, &p3.x, &p3.y, &p4.x, &p4.y);
// 平行
if (isParallel(p1, p2, p3, p4))
printf("2\n");
// 正交
else if (isOrthogonal(p1, p2, p3, p4))
printf("1\n");
else
printf("0\n");
}
输出各点在直线p1p2的投影
Segment s;
Point p, ans;
int n;
scanf("%lf %lf %lf %lf", &s.p1.x, &s.p1.y, &s.p2.x, &s.p2.y);
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &p.x, &p.y);
ans = project(s, p);
printf("%.10lf %.10lf\n", ans.x, ans.y);
}
求点在这条直线的对称点
Segment s;
Point p, ans;
int n;
scanf("%lf %lf %lf %lf", &s.p1.x, &s.p1.y, &s.p2.x, &s.p2.y);
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &p.x, &p.y);
ans = reflect(s, p);
printf("%.10lf %.10lf\n", ans.x, ans.y);
}
求线段s1和s2之间的距离
Segment s1, s2;
int n;
scanf("%d", &n);
while (n--) {
scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &s1.p1.x, &s1.p1.y,
&s1.p2.x, &s1.p2.y, &s2.p1.x, &s2.p1.y,
&s2.p2.x, &s2.p2.y);
printf("%.10lf\n", getDistance(s1, s2));
}
求三个点的逆时针方向
Point p1, p2;
Point p, ans;
int n;
scanf("%lf %lf %lf %lf", &p1.x, &p1.y, &p2.x, &p2.y);
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &p.x, &p.y);
int ans = ccw(p1, p2, p);
if (ans == 1)
printf("COUNTER_CLOCKWISE\n");
else if (ans == -1)
printf("CLOCKWISE\n");
else if (ans == 2)
printf("ONLINE_BACK\n");
else if (ans == -2)
printf("ONLINE_FRONT\n");
else
printf("ON_SEGMENT\n");
}
判断两个线段是否相交
Segment s1, s2;
int n;
scanf("%d", &n);
while (n--) {
scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &s1.p1.x, &s1.p1.y,
&s1.p2.x, &s1.p2.y, &s2.p1.x, &s2.p1.y,
&s2.p2.x, &s2.p2.y);
printf("%d\n", intersect(s1, s2) ? 1 : 0);
}
求线段s1,s2的交点
Segment s1, s2;
int n;
scanf("%d", &n);
while (n--) {
scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &s1.p1.x, &s1.p1.y,
&s1.p2.x, &s1.p2.y, &s2.p1.x, &s2.p1.y,
&s2.p2.x, &s2.p2.y);
Point ans = getCrossPoint(s1, s2);
printf("%.10lf %.10lf\n", ans.x, ans.y);
}
CGL_7_D:Cross Points of a Circle and a Line
求圆和直线的交点,题目限制圆与直线存在交点。实际上,只要检查圆心到直线的距离是否大于r就知道圆与直线是否存在交点。
Circle c;
Line l;
int n;
scanf("%lf %lf %lf", &c.c.x, &c.c.y, &c.r);
scanf("%d", &n);
while (n--) {
scanf("%lf %lf %lf %lf", &l.p1.x, &l.p1.y,
&l.p2.x, &l.p2.y);
pair<Point, Point> ans = getCrossPoints(c, l);
if(ans.first < ans.second)
printf("%.10lf %.10lf %.10lf %.10lf\n", ans.first.x, ans.first.y,
ans.second.x, ans.second.y);
else
printf("%.10lf %.10lf %.10lf %.10lf\n", ans.second.x, ans.second.y,
ans.first.x, ans.first.y);
}
CGL_7_E:Cross Points of Circles
求两个圆的交点,题目限制两个圆存在交点。实际上,只要检查两圆心的距离是否大于r1 + r2就知道两个圆是否存在交点。
Circle c1, c2;
scanf("%lf %lf %lf", &c1.c.x, &c1.c.y, &c1.r);
scanf("%lf %lf %lf", &c2.c.x, &c2.c.y, &c2.r);
pair<Point, Point> ans = getCrossPoints(c1, c2);
if (ans.first < ans.second)
printf("%.10lf %.10lf %.10lf %.10lf\n", ans.first.x, ans.first.y,
ans.second.x, ans.second.y);
else
printf("%.10lf %.10lf %.10lf %.10lf\n", ans.second.x, ans.second.y,
ans.first.x, ans.first.y);
CGL_3_C:Polygon-Point Containment
求点在多边形的位置,若在里面则输出2,若在边上则输出1,在外面则输出0
Polygon g;
Point p ,t;
int n, q;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &t.x, &t.y);
g.push_back(t);
}
scanf("%d", &q);
for (int i = 0; i < q; i++) {
scanf("%lf %lf", &p.x, &p.y);
printf("%d\n", contains(g, p));
}
求点集合P的凸包,需要按照从最左侧最下端开始输出。
Polygon g, ans ,temp;
Point t;
int n, len;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lf %lf", &t.x, &t.y);
g.push_back(t);
}
ans = andrewScan(g);
temp = ans;
// 选出最左侧最下端的点开始输出
sort(temp.begin(), temp.end(),cmp);
int index;
for (int i = 0; i < ans.size(); i++)
if (ans[i] == temp[0])
index = i;
printf("%d\n", ans.size());
for (int i = 0; i < ans.size(); i++)
printf("%d %d\n", int(ans[(i + index) % ans.size()].x),
int(ans[(i + index) % ans.size()].y));