标签：return polygon Point double sgn Scrambled 极角 convex POJ2007
A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the vertices of the polygon. When one starts at any vertex of a closed polygon and traverses each bounding line segment exactly once, one comes back to the starting vertex.A closed polygon is called convex if the line segment joining any two points of the polygon lies in the polygon. Figure 1 shows a closed polygon which is convex and one which is not convex. (Informally, a closed polygon is convex if its border doesn't have any "dents".)
The subject of this problem is a closed convex polygon in the coordinate plane, one of whose vertices is the origin (x = 0, y = 0). Figure 2 shows an example. Such a polygon will have two properties significant for this problem.
The first property is that the vertices of the polygon will be confined to three or fewer of the four quadrants of the coordinate plane. In the example shown in Figure 2, none of the vertices are in the second quadrant (where x < 0, y > 0).
To describe the second property, suppose you "take a trip" around the polygon: start at (0, 0), visit all other vertices exactly once, and arrive at (0, 0). As you visit each vertex (other than (0, 0)), draw the diagonal that connects the current vertex with (0, 0), and calculate the slope of this diagonal. Then, within each quadrant, the slopes of these diagonals will form a decreasing or increasing sequence of numbers, i.e., they will be sorted. Figure 3 illustrates this point.
Input
The input lists the vertices of a closed convex polygon in the plane. The number of lines in the input will be at least three but no more than 50. Each line contains the x and y coordinates of one vertex. Each x and y coordinate is an integer in the range -999..999. The vertex on the first line of the input file will be the origin, i.e., x = 0 and y = 0. Otherwise, the vertices may be in a scrambled order. Except for the origin, no vertex will be on the x-axis or the y-axis. No three vertices are colinear.Output
输出列出给定多边形的顶点，每行一个顶点。输入的每个顶点在输出中正好显示一次。原点 （0，0） 是输出第一行的顶点。输出中的紫杉顺序将决定沿多边形边界沿逆时针方向行驶。每个顶点的输出格式如下所示。Sample Input
0 0
70 -50
60 30
-30 -50
80 20
50 -60
90 -20
-30 -40
-10 -60
90 10
Sample Output
(0,0)
(-30,-40)
(-30,-50)
(-10,-60)
(50,-60)
(70,-50)
(90,-20)
(90,10)
(80,20)
(60,30)题意： 给出一个凸多边形的n个点，且其一定包含原点，从原点开始逆时针输出凸多边形上的点。
分析： 由于已知所有点构成凸多边形就不需要求凸包了，直接对极角排序。顺便介绍一下常用的两种极角排序方法，第一种是调用cmath库函数atan2(double y, double x)，返回值就是点(x, y)的弧度极角，取值范围[-pi, pi]，因此可以直接对该值进行排序。第二种方法是利用叉积，先把原点选定为最左下角的点，cmp中return aXb大于0。
排完序以后起点可能不是零点，因此输出时要先找到零点，从零点开始输出。
具体代码如下：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <cstring>
using namespace std;
// `计算几何模板`
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}
/*
* Point
* Point() - Empty constructor
* Point(double _x,double _y) - constructor
* input() - double input
* output() - %.2f output
* operator == - compares x and y
* operator < - compares first by x, then by y
* operator - - return new Point after subtracting curresponging x and y
* operator ^ - cross product of 2d points
* operator * - dot product
* len() - gives length from origin
* len2() - gives square of length from origin
* distance(Point p) - gives distance from p
* operator + Point b - returns new Point after adding curresponging x and y
* operator * double k - returns new Point after multiplieing x and y by k
* operator / double k - returns new Point after divideing x and y by k
* rad(Point a,Point b)- returns the angle of Point a and Point b from this Point
* trunc(double r) - return Point that if truncated the distance from center to r
* rotleft() - returns 90 degree ccw rotated point
* rotright() - returns 90 degree cw rotated point
* rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw
*/
struct Point
{
double x, y;
Point(){}
Point(double _x,double _y){x = _x, y = _y;}
void input(){scanf("%lf%lf",&x,&y);}
void output(){printf("%.2f %.2f\n",x,y);}
bool operator == (Point b)const{return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;}
bool operator < (Point b)const{return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;}
Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}
//叉积
double operator ^(const Point &b)const{return x*b.y - y*b.x;}
//点积
double operator *(const Point &b)const{return x*b.x + y*b.y;}
//返回长度
double len(){return hypot(x,y);/*库函数*/}
//返回长度的平方
double len2(){return x*x + y*y;}
//返回两点的距离
double distance(Point p){return hypot(x-p.x,y-p.y);}
Point operator +(const Point &b)const{return Point(x+b.x,y+b.y);}
Point operator *(const double &k)const{return Point(x*k,y*k);}
Point operator /(const double &k)const{return Point(x/k,y/k);}
//`计算pa 和 pb 的夹角`
//`就是求这个点看a,b 所成的夹角`
//`测试 LightOJ1203`
double rad(Point a,Point b)
{
Point p = *this;
return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
}
//`化为长度为r的向量`
Point trunc(double r)
{
double l = len();
if(!sgn(l))return *this;
r /= l;
return Point(x*r,y*r);
}
//`逆时针旋转90度`
Point rotleft(){return Point(-y,x);}
//`顺时针旋转90度`
Point rotright(){return Point(y,-x);}
//`绕着p点逆时针旋转angle`
Point rotate(Point p,double angle)
{
Point v = (*this) - p;
double c = cos(angle), s = sin(angle);
return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
}
}p[105];
/*
* Stores two points
* Line() - Empty constructor
* Line(Point _s,Point _e) - Line through _s and _e
* operator == - checks if two points are same
* Line(Point p,double angle) - one end p , another end at angle degree
* Line(double a,double b,double c) - Line of equation ax + by + c = 0
* input() - inputs s and e
* adjust() - orders in such a way that s < e
* length() - distance of se
* angle() - return 0 <= angle < pi
* relation(Point p) - 3 if point is on line
* 1 if point on the left of line
* 2 if point on the right of line
* pointonseg(double p) - return true if point on segment
* parallel(Line v) - return true if they are parallel
* segcrossseg(Line v) - returns 0 if does not intersect
* returns 1 if non-standard intersection
* returns 2 if intersects
* linecrossseg(Line v) - line and seg
* linecrossline(Line v) - 0 if parallel
* 1 if coincides
* 2 if intersects
* crosspoint(Line v) - returns intersection point
* dispointtoline(Point p) - distance from point p to the line
* dispointtoseg(Point p) - distance from p to the segment
* dissegtoseg(Line v) - distance of two segment
* lineprog(Point p) - returns projected point p on se line
* symmetrypoint(Point p) - returns reflection point of p over se
*
*/
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e){s = _s, e = _e;}
bool operator ==(Line v){return (s == v.s)&&(e == v.e);}
//`根据一个点和倾斜角angle确定直线,0<=angle<pi`
Line(Point p,double angle)
{
s = p;
if(sgn(angle-pi/2) == 0){e = (s + Point(0,1));}
else{e = (s + Point(1,tan(angle)));}
}
//ax+by+c=0
Line(double a,double b,double c)
{
if(sgn(a) == 0) s = Point(0,-c/b), e = Point(1,-c/b);
else if(sgn(b) == 0) s = Point(-c/a,0), e = Point(-c/a,1);
else s = Point(0,-c/b), e = Point(1,(-c-a)/b);
}
void input()
{
s.input();
e.input();
}
void adjust(){if(e < s)swap(s,e);}
//求线段长度
double length(){return s.distance(e);}
//`返回直线倾斜角 0<=angle<pi`
double angle()
{
double k = atan2(e.y-s.y,e.x-s.x);
if(sgn(k) < 0)k += pi;
if(sgn(k-pi) == 0)k -= pi;
return k;
}
//`点和直线关系`
//`1 在左侧`
//`2 在右侧`
//`3 在直线上`
int relation(Point p)
{
int c = sgn((p-s)^(e-s));
if(c < 0)return 1;
else if(c > 0)return 2;
else return 3;
}
// 点在线段上的判断
bool pointonseg(Point p){return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;}
//`两向量平行(对应直线平行或重合)`
bool parallel(Line v){return sgn((e-s)^(v.e-v.s)) == 0;/*两向量叉积为0*/ }
//`两线段相交判断`
//`2 规范相交`
//`1 非规范相交`
//`0 不相交`
int segcrossseg(Line v)
{
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
int d3 = sgn((v.e-v.s)^(s-v.s));
int d4 = sgn((v.e-v.s)^(e-v.s));
if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
}
//`直线和线段相交判断`
//`-*this line -v seg`
//`2 规范相交`
//`1 非规范相交`
//`0 不相交`
int linecrossseg(Line v)
{
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
if((d1^d2)==-2) return 2;
return (d1==0||d2==0);
}
//`两直线关系`
//`0 平行`
//`1 重合`
//`2 相交`
int linecrossline(Line v)
{
if((*this).parallel(v))//此时平行或者重合
return v.relation(s)==3;//如果当前直线起点在另一条直线上
return 2;
}
//`求两直线的交点`
//`要保证两直线不平行或重合`
Point crosspoint(Line v)
{
double a1 = (v.e-v.s)^(s-v.s);
double a2 = (v.e-v.s)^(e-v.s);
return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
}
//点到直线的距离
double dispointtoline(Point p){return fabs((p-s)^(e-s))/length();}
//点到线段的距离
double dispointtoseg(Point p)
{
if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
return min(p.distance(s),p.distance(e));
return dispointtoline(p);
}
//`返回线段到线段的距离`
//`前提是两线段不相交，相交距离就是0了`
double dissegtoseg(Line v){return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));}
//`返回点p在直线上的投影`
Point lineprog(Point p){return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );}
//`返回点p关于直线的对称点`
Point symmetrypoint(Point p)
{
Point q = lineprog(p);
return Point(2*q.x-p.x,2*q.y-p.y);
}
};
struct polygon{
int n;
Point p[maxp];
Line l[maxp];
void input(int _n){
n = _n;
for(int i = 0;i < n;i++)
p[i].input();
}
void add(Point q){
p[n++] = q;
}
void getline(){
for(int i = 0;i < n;i++){
l[i] = Line(p[i],p[(i+1)%n]);
}
}
struct cmp{
Point p;
cmp(const Point &p0){p = p0;}
bool operator()(const Point &aa,const Point &bb){
Point a = aa, b = bb;
int d = sgn((a-p)^(b-p));
if(d == 0){
return sgn(a.distance(p)-b.distance(p)) < 0;
}
return d > 0;
}
};
//`进行极角排序`
//`首先需要找到最左下角的点`
//`需要重载号好Point的 < 操作符(min函数要用) `
void norm(){
Point mi = p[0];
for(int i = 1;i < n;i++)mi = min(mi,p[i]);
sort(p,p+n,cmp(mi));
}
//`得到凸包`
//`得到的凸包里面的点编号是0$\sim$n-1的`
//`两种凸包的方法`
//`注意如果有影响，要特判下所有点共点，或者共线的特殊情况`
//`测试 LightOJ1203 LightOJ1239`
void getconvex(polygon &convex){
sort(p,p+n);
convex.n = n;
for(int i = 0;i < min(n,2);i++){
convex.p[i] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
if(n <= 2)return;
int &top = convex.n;
top = 1;
for(int i = 2;i < n;i++){
while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)
top--;
convex.p[++top] = p[i];
}
int temp = top;
convex.p[++top] = p[n-2];
for(int i = n-3;i >= 0;i--){
while(top != temp && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)
top--;
convex.p[++top] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
convex.norm();//`原来得到的是顺时针的点，排序后逆时针`
}
//`得到凸包的另外一种方法`
//`测试 LightOJ1203 LightOJ1239`
void Graham(polygon &convex){
norm();
int &top = convex.n;
top = 0;
if(n == 1){
top = 1;
convex.p[0] = p[0];
return;
}
if(n == 2){
top = 2;
convex.p[0] = p[0];
convex.p[1] = p[1];
if(convex.p[0] == convex.p[1])top--;
return;
}
convex.p[0] = p[0];
convex.p[1] = p[1];
top = 2;
for(int i = 2;i < n;i++){
while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2])) <= 0 )
top--;
convex.p[top++] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
}
//`判断是不是凸的`
bool isconvex(){
bool s[2];
memset(s,false,sizeof(s));
for(int i = 0;i < n;i++){
int j = (i+1)%n;
int k = (j+1)%n;
s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;
if(s[0] && s[2])return false;
}
return true;
}
//`判断点和任意多边形的关系`
//` 3 点上`
//` 2 边上`
//` 1 内部`
//` 0 外部`
int relationpoint(Point q){
for(int i = 0;i < n;i++){
if(p[i] == q)return 3;
}
getline();
for(int i = 0;i < n;i++){
if(l[i].pointonseg(q))return 2;
}
int cnt = 0;
for(int i = 0;i < n;i++){
int j = (i+1)%n;
int k = sgn((q-p[j])^(p[i]-p[j]));
int u = sgn(p[i].y-q.y);
int v = sgn(p[j].y-q.y);
if(k > 0 && u < 0 && v >= 0)cnt++;
if(k < 0 && v < 0 && u >= 0)cnt--;
}
return cnt != 0;
}
//`直线u切割凸多边形左侧`
//`注意直线方向`
//`测试：HDU3982`
void convexcut(Line u,polygon &po){
int &top = po.n;//注意引用
top = 0;
for(int i = 0;i < n;i++){
int d1 = sgn((u.e-u.s)^(p[i]-u.s));
int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));
if(d1 >= 0)po.p[top++] = p[i];
if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i+1)%n]));
}
}
//`得到周长`
//`测试 LightOJ1239`
double getcircumference(){
double sum = 0;
for(int i = 0;i < n;i++){
sum += p[i].distance(p[(i+1)%n]);
}
return sum;
}
//`得到面积`
double getarea(){
double sum = 0;
for(int i = 0;i < n;i++){
sum += (p[i]^p[(i+1)%n]);
}
return fabs(sum)/2;
}
//`得到方向`
//` 1 表示逆时针，0表示顺时针`
bool getdir(){
double sum = 0;
for(int i = 0;i < n;i++)
sum += (p[i]^p[(i+1)%n]);
if(sgn(sum) > 0)return 1;
return 0;
}
//`得到重心`
Point getbarycentre(){
Point ret(0,0);
double area = 0;
for(int i = 1;i < n-1;i++){
double tmp = (p[i]-p[0])^(p[i+1]-p[0]);
if(sgn(tmp) == 0)continue;
area += tmp;
ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;
ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;
}
if(sgn(area)) ret = ret/area;
return ret;
}
};
signed main()
{
int cnt = 0;
double tx, ty;
while(~scanf("%lf%lf", &tx, &ty))
{
p[++cnt].x = tx;
p[cnt].y = ty;
}
polygon a;
a.n = cnt;
for(int i = 1; i <= cnt; i++)
a.p[i-1] = p[i];
a.norm();
for(int i = 0; i < cnt; i++)
{
if(!sgn(a.p[i].x) && !sgn(a.p[i].y))
{
for(int j = i; j < cnt; j++)
printf("(%.0f,%.0f)\n", a.p[j].x, a.p[j].y);
for(int j = 0; j < i; j++)
printf("(%.0f,%.0f)\n", a.p[j].x, a.p[j].y);
return 0;
}
}
return 0;
}标签：return,polygon,Point,double,sgn,Scrambled,极角,convex,POJ2007 来源： https://blog.csdn.net/m0_55982600/article/details/120975217
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