POJ:1113
2023-12-13 12:37:14
题目链接:http://poj.org/problem?id=1113?
求凸包周长
因为拐角处的围墙是可以弯曲的,根据多边形内角和可以得出弯的城墙长度刚好是一个圆
故ans=2*pi*l+C
/*
floor();//向下取整 ceil();//向上取整 round();//四舍五入
判等时不要直接使用=
*/
#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1.0);
const double inf = 1e100;
/*
误差判断
*/
const double eps = 1e-9;
int dcmp(double x, double y){
if(fabs(x - y) < eps)
return 0;
if(x > y)
return 1;
return -1;
}
int sgn(double d){
if(fabs(d) < eps)
return 0;
if(d > 0)
return 1;
return -1;
}
/*---------------------------------------------------------------------------------------*/
//点
struct Point{
double x, y,z;
Point(double x = 0, double y = 0, double z = 0):x(x),y(y),z(z){}
};
//向量
typedef Point Vector;
//运算
Vector operator + (Vector A, Vector B){
return Vector(A.x+B.x, A.y+B.y,A.z+B.z);
}
Vector operator - (Vector A, Vector B){
return Vector(A.x-B.x, A.y-B.y,A.z-B.z);
}
Vector operator * (Vector A, double p){
return Vector(A.x*p, A.y*p,A.z*p);
}
Vector operator / (Vector A, double p){
return Vector(A.x/p, A.y/p,A.z/p);
}
bool operator == (Point A,Point B){
return sgn(A.x-B.x)==0&&sgn(A.y-B.y)==0;
}
bool cmp1(Point a,Point b){
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
//向量数量积 向量α在向量β的投影于向量β的长度乘积(带方向)
double Dot(Vector A, Vector B){
return A.x*B.x + A.y*B.y + A.z*B.z;
}
//向量向量积 向量α与β所张成的平行四边形的有向面积
double Cross(Vector A, Vector B){
return A.x*B.y-A.y*B.x;
}
//取模
double Length(Vector A){
return sqrt(Dot(A, A));
}
//计算夹角
double Angle(Vector A, Vector B){
return acos(Dot(A, B) / Length(A) / Length(B));
}
//向量旋转
Vector Rotate(Vector A, double rad){//rad为弧度 且为逆时针旋转的角
return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
Vector Normal(Vector A){//向量A左转90°的单位法向量
double L = Length(A);
return Vector(-A.y/L, A.x/L);
}
//三角形外接圆圆心
Point Excenter(Point a, Point b, Point c){
double a1 = b.x - a.x;
double b1 = b.y - a.y;
double c1 = (a1*a1 + b1*b1)/2;
double a2 = c.x - a.x;
double b2 = c.y - a.y;
double c2 = (a2*a2 + b2*b2)/2;
double d = a1*b2 - a2*b1;
return Point(a.x + (c1*b2 - c2*b1)/d, a.y + (a1*c2 - a2*c1)/d);
}
//两点距离
double dis(Point a,Point b){
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
/*------------------------------------------------------------------------------------------------*/
//线
struct Line{//直线定义
Point v, p;
Line(Point v, Point p):v(v), p(p) {}
Point point(double t){//返回点P = v + (p - v)*t
return v + (p - v)*t;
}
};
//判断点是否在线上
bool PointOnline(Line A,Point a){
Vector v1(A.p.x-a.x,A.p.y-a.y,A.p.z-a.z);
Vector v2(A.v.x-a.x,A.v.y-a.y,A.v.z-a.z);
return !(Cross(v1,v2));
}
//计算两直线交点 必须保证直线相交,否则将会出现除以零的情况
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;
}
//点P到直线AB距离公式
double DistanceToLine(Point P, Point A, Point B){
Vector v1 = B-A, v2 = P-A;
return fabs(Cross(v1, v2)/Length(v1));
}
//点P到线段AB距离公式
double DistanceToSegment(Point P, Point A, Point B){
if(A.x == B.x&&A.y == B.y&&A.z == B.z)
return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2),0) < 0)
return Length(v2);
if(dcmp(Dot(v1, v3),0) > 0)
return Length(v3);
return DistanceToLine(P, A, B);
}
//点P在直线AB上的投影点
Point GetLineProjection(Point P, Point A, Point B){
Vector v = B-A;
return A+v*(Dot(v, P-A)/Dot(v, v));
}
//判断点是否在线段端上
bool OnSegment(Point p, Point a1, Point a2){
return dcmp(Cross(a1-p, a2-p),0) == 0 && dcmp(Dot(a1-p, a2-p),0) <= 0;
}
//判断线段相交
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){
double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);
double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
//if判断控制是否允许线段在端点处相交,根据需要添加
if(!sgn(c1) || !sgn(c2) || !sgn(c3) || !sgn(c4)){
bool f1 = OnSegment(b1, a1, a2);
bool f2 = OnSegment(b2, a1, a2);
bool f3 = OnSegment(a1, b1, b2);
bool f4 = OnSegment(a2, b1, b2);
bool f = (f1|f2|f3|f4);
return f;
}
return (sgn(c1)*sgn(c2) < 0 && sgn(c3)*sgn(c4) < 0);
}
//点c是否在线段ab的左侧
bool ToLeftTest(Point a, Point b, Point c){
return Cross(b - a, c - b) > 0;
}
/*------------------------------------------------------------------------*/
//多边形
//求凸包 多边形p 凸包ch
int ConvexHull(Point *p,int n,Point *ch){
n=unique(p,p+n)-p;//去重
sort(p,p+n,cmp1);
int v=0;
for(int i=0;i<n;i++){
while(v>1&&Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1])<=0)
v--;
ch[v++]=p[i];
}
int j=v;
for(int i=n-2;i>=0;i--){
while(v>j&&Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1])<=0)
v--;
ch[v++]=p[i];
}
if(n>1) v--;
return v;
}
//多边形有向面积
double PolygonArea(Point* p, int n){//p为端点集合,n为端点个数
double s = 0;
for(int i = 1; i < n-1; ++i)
s += Cross(p[i]-p[0], p[i+1]-p[0]);
return s;
}
//判断点是否在多边形内,若点在多边形内返回1,在多边形外部返回0,在多边形上返回-1
int isPointInPolygon(Point p, vector<Point> poly){
int wn = 0;
int n = poly.size();
for(int i = 0; i < n; ++i){
if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1;
int k = sgn(Cross(poly[(i+1)%n] - poly[i], p - poly[i]));
int d1 = sgn(poly[i].y - p.y);
int d2 = sgn(poly[(i+1)%n].y - p.y);
if(k > 0 && d1 <= 0 && d2 > 0) wn++;
if(k < 0 && d2 <= 0 && d1 > 0) wn--;
}
if(wn != 0)
return 1;
return 0;
}
#define pi 3.1415926535
int main()
{
Point p[1003],ch[1003];
int n,l;
cin>>n>>l;
for(int i=0;i<n;i++)
cin>>p[i].x>>p[i].y;
int m=ConvexHull(p,n,ch);
double ans=0;
for(int i=0;i<m;i++)
ans+=dis(ch[i],ch[(i+1)%m]);
ans+=(2.0*pi*l);
printf("%.0lf\n",ans);
}
文章来源:https://blog.csdn.net/weixin_63238732/article/details/132737059
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