QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#638986 | #6599. The Grand Tournament | asitshouldbe | WA | 245ms | 4356kb | C++17 | 30.7kb | 2024-10-13 17:31:15 | 2024-10-13 17:31:15 |
Judging History
answer
#include <bits/stdc++.h>
#define pi acos(-1.0)
//#define double long double
using namespace std;
const double eps = 1e-9,inf=1e12;
const int N=1e3+10;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
if(x < 0) return -1;
else return 1;
}
int dsgn(double x,double y)
{
if(fabs(x-y)<eps) return 0;
if(x<y) return -1;
else return 1;
}
inline double sqr(double x){return x*x;}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y){x = _x;y = _y;}
void input(){cin>>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);}
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;}
//数乘
Point operator *(const double &k)const{
return Point(x*k,y*k);
}
Point operator /(const double &k)const{
return Point(x/k,y/k);
}
//返回长度
double len(){return hypot(x,y);}
//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);}
int distance2(Point p){return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);}
//`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) ));
}
//`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);
}
//`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);}
};
//`AB X AC`
double cross(Point A,Point B,Point C){return (B-A)^(C-A);}
//`AB*AC`
double dot(Point A,Point B,Point C){return (B-A)*(C-A);}
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;
}
bool operator <(Line& v){
return sgn(angle()-v.angle())==0?((e-s)^(v.e-s))<0:angle()<v.angle();
}
//`点和直线关系`
int relation(Point p){
int c = sgn((p-s)^(e-s));
if(c < 0)return 1; //`1 在左侧`
else if(c > 0)return 2; //`2 在右侧`
else return 3; //`3 在直线上`
}
//`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;
}
//`两向量正交`
bool orthogonal(Line v){
return sgn((e-s)*(v.e-v.s)) == 0;
}
//`两直线关系`
int linecrossline(Line v){
if((*this).parallel(v)) //`0 平行`
return v.relation(s)==3; //`1 重合`
if((*this).orthogonal(v)) return 3; //`3 正交`
else return 2; //`2 相交`
}
//`两线段相交判断`
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; //`2 规范相交`
return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) || //`1 非规范相交`
(d2==0 && sgn((v.e-s)*(v.e-e))<=0) || //`0 不相交`
(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
}
int check(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 0;
bool f=(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);
if(!f) return 0;
if(e==v.e&&s==v.s) return 1; //完全重合
if(e==v.e||e==v.s||s==v.e||s==v.s)
{
if((*this).parallel(v))
{
if(e==v.e&&((*this).pointonseg(v.s)||v.pointonseg(s))) return 3;//端点重合部分重合
if(e==v.s&&((*this).pointonseg(v.e)||v.pointonseg(s))) return 3;
if(s==v.e&&((*this).pointonseg(v.s)||v.pointonseg(e))) return 3;
if(s==v.s&&((*this).pointonseg(v.e)||v.pointonseg(e))) return 3;
else return 0;
}
return 2; //端点重合
}
int a1=(*this).pointonseg(v.s);
int a2=(*this).pointonseg(v.e);
int a3=v.pointonseg(e);
int a4=v.pointonseg(s);
if((a1||a2)&&(a3||a4)) return 4;
if(a1&&a2) return 4;
if(a3&&a4) return 4;
return 0;
// if(a1+a2+a3+a4==1) return 0;
// return 4;
}
//`直线和线段相交判断`
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; //`2 规范相交`
return (d1==0||d2==0); //`1 非规范相交 0 不相交`
}
//`求两直线的交点 要保证两直线不平行或重合`
Point crosspoint(Line l)
{
Point u=e-s,v=l.e-l.s;
double t=(s-l.s)^v/(v^u);
return s+u*t;
}
//`返回点p在直线上的投影`
Point lineprog(Point p){
Point u=e-s,v=p-s;
return s + u*(u*v/u.len2());
}
//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关于直线的对称点`
Point symmetrypoint(Point p){
Point pro = lineprog(p);
return pro*2-p;
}
};
struct circle{
Point p;//圆心
double r;//半径
circle(){}
circle(Point _p,double _r){p = _p;r = _r;}
circle(double x,double y,double _r){p = Point(x,y);r = _r;}
//`三角形的外接圆`
//`需要Point的+ / rotate() 以及Line的crosspoint()`
//`利用两条边的中垂线得到圆心`
//`测试:UVA12304`
circle(Point a,Point b,Point c){
Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
p = u.crosspoint(v);
r = p.distance(a);
}
//`三角形的内切圆`
//`参数bool t没有作用,只是为了和上面外接圆函数区别`
//`测试:UVA12304`
circle(Point a,Point b,Point c,bool t){
Line u,v;
double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x);
u.s = a;
u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
v.s = b;
m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
p = u.crosspoint(v);
r = Line(a,b).dispointtoseg(p);
}
//输入
void input(){p.input();cin>>r;}
bool operator == (circle v){
return (p==v.p) && sgn(r-v.r)==0;
}
bool operator < (circle v)const{
return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
}
//面积
double area(){return pi*r*r;}
//周长
double circumference(){return 2*pi*r;}
//`点和圆的关系`
int relation(Point b){
double dst = b.distance(p);
if(sgn(dst-r) < 0)return 2; //`2 圆内`
else if(sgn(dst-r)==0)return 1; //`1 圆上`
return 0; //`0 圆外`
}
//`线段和圆的关系 比较的是圆心到线段的距离和半径的关系`
int relationseg(Line v){
double dst = v.dispointtoseg(p);
if(sgn(dst-r) < 0)return 2; //`2 相交`
else if(sgn(dst-r) == 0)return 1; //`1 相切`
return 0; //`0 相离`
}
//`直线和圆的关系 比较的是圆心到直线的距离和半径的关系`
int relationline(Line v){
double dst = v.dispointtoline(p);
if(sgn(dst-r) < 0)return 2; //`2 相交`
else if(sgn(dst-r) == 0)return 1; //`1 相切`
return 0; //`0 相离`
}
//`两圆的关系 需要Point的distance`
//`测试:UVA12304`
int relationcircle(circle v){
double d = p.distance(v.p);
if(sgn(d-r-v.r) > 0)return 5; //`5 相离`
if(sgn(d-r-v.r) == 0)return 4; //`4 外切`
double l = fabs(r-v.r);
if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3; //`3 相交`
if(sgn(d-l)==0)return 2; //`2 内切`
if(sgn(d-l)<0)return 1; //`1 内含`
}
//`求两个圆的交点 需要relationcircle`
//`测试:UVA12304`
int pointcrosscircle(circle v,Point &p1,Point &p2){
int rel = relationcircle(v);
if(rel == 1 || rel == 5)return 0; //`0 无交点`
double d = p.distance(v.p);
double l = (d*d+r*r-v.r*v.r)/(2*d);
double h = sqrt(r*r-l*l);
Point tmp = p + (v.p-p).trunc(l);
p1 = tmp + ((v.p-p).rotleft().trunc(h));
p2 = tmp + ((v.p-p).rotright().trunc(h));
if(rel == 2 || rel == 4) return 1; //`1 一个交点`
return 2; //`2 两个交点`
}
//`求直线和圆的交点,返回交点个数`
int pointcrossline(Line v,Point &p1,Point &p2){
if(!(*this).relationline(v))return 0;
Point a = v.lineprog(p);
double d = v.dispointtoline(p);
d = sqrt(r*r-d*d);
if(sgn(d) == 0){
p1 = a;
p2 = a;
return 1;
}
p1 = a + (v.e-v.s).trunc(d);
p2 = a - (v.e-v.s).trunc(d);
return 2;
}
//`得到过a,b两点,半径为r1的两个圆`
int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){
circle x(a,r1),y(b,r1);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r;
return t; //`返回圆的个数`
}
//`得到与直线u相切,过点q,半径为r1的圆`
//`测试:UVA12304`
int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){
double dis = u.dispointtoline(q);
if(sgn(dis-r1*2)>0)return 0;
if(sgn(dis) == 0){
c1.p = q + ((u.e-u.s).rotleft().trunc(r1));
c2.p = q + ((u.e-u.s).rotright().trunc(r1));
c1.r = c2.r = r1;
return 2;
}
Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e + (u.e-u.s).rotleft().trunc(r1)));
Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e + (u.e-u.s).rotright().trunc(r1)));
circle cc = circle(q,r1);
Point p1,p2;
if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2);
c1 = circle(p1,r1);
if(p1 == p2){
c2 = c1;
return 1;
}
c2 = circle(p2,r1);
return 2;
}
//`同时与直线u,v相切,半径为r1的圆`
//`测试:UVA12304`
int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4){
if(u.parallel(v))return 0;//两直线平行
Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u.e-u.s).rotleft().trunc(r1));
Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (u.e-u.s).rotright().trunc(r1));
Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v.e-v.s).rotleft().trunc(r1));
Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (v.e-v.s).rotright().trunc(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = u1.crosspoint(v1);
c2.p = u1.crosspoint(v2);
c3.p = u2.crosspoint(v1);
c4.p = u2.crosspoint(v2);
return 4;
}
//`同时与不相交圆cx,cy相切,半径为r1的圆`
//`测试:UVA12304`
int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2){
circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r1;
return t; //`返回圆的个数`
}
//`过一点作圆的切线(先判断点和圆的关系)`
//`测试:UVA12304`
int tangentline(Point q,Line &u,Line &v){
int x = relation(q);
if(x == 2)return 0;
if(x == 1){
u = Line(q,q + (q-p).rotleft());
v = u;
return 1;
}
double d = p.distance(q);
double l = r*r/d;
double h = sqrt(r*r-l*l);
u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)));
v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h)));
return 2; //`返回切线的个数`
}
int tangentpoint(Point q,Point &u,Point &v){
int x = relation(q);
if(x == 2)return 0;
if(x == 1){
u = q;v = u;
return 1;
}
double d = p.distance(q);
double l = r*r/d;
double h = sqrt(r*r-l*l);
u = p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h));
v = p + ((q-p).trunc(l) + (q-p).rotright().trunc(h));
return 2; //`返回切点的个数`
}
//`求两圆相交的面积`
double areacircle(circle v){
int rel = relationcircle(v);
if(rel >= 4)return 0.0;
if(rel <= 2)return min(area(),v.area());
double d = p.distance(v.p);
double hf = (r+v.r+d)/2.0;
double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
a1 = a1*r*r;
double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
a2 = a2*v.r*v.r;
return a1+a2-ss;
}
//`求圆和三角形pab的相交面积`
//`测试:POJ3675 HDU3982 HDU2892`
double areatriangle(Point a,Point b){
if(sgn((p-a)^(p-b)) == 0)return 0.0;
Point q[5];
int len = 0;
q[len++] = a;
Line l(a,b);
Point p1,p2;
if(pointcrossline(l,q[1],q[2])==2){
if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];
if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];
}
q[len++] = b;
if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q[2]);
double res = 0;
for(int i = 0;i < len-1;i++){
if(relation(q[i])==0||relation(q[i+1])==0){
double arg = p.rad(q[i],q[i+1]);
res += r*r*arg/2.0;
}
else res += fabs((q[i]-p)^(q[i+1]-p))/2.0;
}
return res;
}
//`两圆公切线`
Point getpoint(double rad){return Point(p.x+r*cos(rad),p.y+r*sin(rad));}
int conmontangent(circle v,vector<Point> &p1,vector<Point> &p2)
{
bool flag=0;
if(r<v.r) swap(*this,v),flag=1;
double d=p.distance(v.p),rd=r-v.r,rs=r+v.r;
if(sgn(d-rd)<0) return 0;
if(sgn(d)==0) return -1;
double rad=Line(p,v.p).angle();
if(sgn(d-rd)==0)
{
p1.push_back(getpoint(rad)),p2.push_back(getpoint(rad));
return 1; //`一条外公切线`
}
double rad1=acos(rd/d);
p1.push_back(getpoint(rad+rad1)),p2.push_back(v.getpoint(rad+rad1));
p1.push_back(getpoint(rad-rad1)),p2.push_back(v.getpoint(rad-rad1));
if(sgn(d-rs)==0)
{
p1.push_back(getpoint(rad)),p2.push_back(getpoint(rad));
if(flag) swap(p1,p2);
return 3; //`两条外公切线 一条内公切线`
}
else if(sgn(d-rs)>0)
{
double rad2=acos(rs/d);
p1.push_back(getpoint(rad+rad2)),p2.push_back(v.getpoint(rad+rad2-pi));
p1.push_back(getpoint(rad-rad2)),p2.push_back(v.getpoint(rad-rad2+pi));
if(flag) swap(p1,p2);
return 4; //`两条外公切线 两条内公切线`
}
else
{
if(flag) swap(p1,p2);
return 2; //`两条外公切线`
}
}
};
struct polygon{
int n;
Point p[N];
Line l[N];
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 mi为极点`
//`需要重载号好Point的 < 操作符(min函数要用) `
void norm(Point mi){
//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-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 andrew(polygon &convex){
sort(p,p+n);
int &top = convex.n;
top = 0;
for(int i = 0;i < n;i++){
while(top>=2&&sgn(cross(convex.p[top-2],convex.p[top-1],p[i]))<=0) top--;
convex.p[top++] = p[i];
}
int temp = top;
for(int i = n-2;i >= 0;i--){
while(top>temp&&sgn(cross(convex.p[top-2],convex.p[top-1],p[i]))<=0) top--;
convex.p[top++] = p[i];
}
top--;
}
//`判断是不是凸的`
bool isconvex(){
bool s[5];
memset(s,false,sizeof(s));
for(int i = 0;i < n;i++){
int j = (i+1)%n,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`
int relationpoint(Point q){
for(int i = 0;i < n;i++){
if(p[i] == q)return 3; //` 3 点上`
}
getline();
for(int i = 0;i < n;i++){
if(l[i].pointonseg(q))return 2; //` 2 边上`
}
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--;
} //` 1 内部`
return cnt != 0; //` 0 外部`
}
//`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`
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; //` 1 逆时针`
else return 0; //` 0 顺时针`
}
//`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;
}
//`多边形和圆交的面积`
//`测试:POJ3675 HDU3982 HDU2892`
double areacircle(circle c){
double ans = 0;
for(int i = 0;i < n;i++){
int j = (i+1)%n;
if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0)
ans += c.areatriangle(p[i],p[j]);
else ans -= c.areatriangle(p[i],p[j]);
}
return fabs(ans);
}
//`多边形和圆关系`
int relationcircle(circle c){
getline();
int x = 2; //` 2 圆完全在多边形内`
if(relationpoint(c.p) != 1)return 0; //` 0 圆心不在内部`
for(int i = 0;i < n;i++){
if(c.relationseg(l[i])==2)return 0; //` 0 其它`
if(c.relationseg(l[i])==1)x = 1; //` 1 圆在多边形里面,碰到了多边形边界`
}
return x;
}
//`旋转卡壳求凸包直径(最远点对)`
double rorating_calipers1()
{
double res=0;
for(int i=0,j=1;i<n;i++)
{
while(dsgn(cross(p[i],p[i+1],p[j]),cross(p[i],p[i+1],p[j+1]))<0) j=(j+1)%n;
res=max(res,max(p[i].distance(p[j]),p[i+1].distance(p[j])));
}
return res;
}
//`旋转卡壳求最小矩形覆盖`
double rorating_calipers2(polygon &pt)
{
double res=1e20;
for(int i=0,a=1,b=1,c;i<n;i++)
{
while(dsgn(cross(p[i],p[i+1],p[a]),cross(p[i],p[i+1],p[a+1]))<0) a=(a+1)%n;
while(dsgn(dot(p[i],p[i+1],p[b]),dot(p[i],p[i+1],p[b+1]))<0) b=(b+1)%n;
if(!i) c=a;
while(dsgn(dot(p[i+1],p[i],p[c]),dot(p[i+1],p[i],p[c+1]))<0) c=(c+1)%n;
double d=p[i].distance(p[i+1]);
double H=cross(p[i],p[i+1],p[a])/d;
double R=dot(p[i],p[i+1],p[b])/d;
double L=dot(p[i+1],p[i],p[c])/d;
if(dsgn(res,(L+R-d)*H)>0)
{
res=(L+R-d)*H;
pt.p[0]=p[i+1]+(p[i]-p[i+1])*(L/d);
pt.p[1]=p[i]+(p[i+1]-p[i])*(R/d);
pt.p[2]=pt.p[1]+(p[i+1]-p[i]).rotleft()*(H/d);
pt.p[3]=pt.p[0]+(p[i+1]-p[i]).rotleft()*(H/d);
}
}
return res;
}
//`分治法求最近点对`
Point a[N];
double divide(int l,int r)
{
if(l==r) return 2e9;
if(l+1==r) return p[l].distance(p[r]);
int mid=l+r>>1;
double d=min(divide(l,mid),divide(mid+1,r));
int k=0;
for(int i=l;i<=r;i++)
if(fabs(p[mid].x-p[i].x)<d) a[k++]=p[i];
//sort(a,a+k,[&](Point a,Point b)->bool {return a.y<b.y;});
for(int i=0;i<k;i++)
for(int j=i+1;j<k&&a[j].y-a[i].y<d;j++)
d=min(d,a[j].distance(a[i]));
return d;
}
//`旋转卡壳求最大三角形面积`
double rotating_calipers3()
{
double res=0;
for(int i=0;i<n;i++)
{
int k=i+1;
for(int j=i+1;j<n;j++)
{
while(dsgn(cross(p[i],p[j],p[k]),cross(p[i],p[j],p[k+1]))<0) k=(k+1)%n;
res=max(res,cross(p[i],p[j],p[k]));
}
}
return res/2;
}
};
//`半平面交求凸多边形面积交`
double half_plane1(Line l[],int n)
{
double res=0;
sort(l,l+n);
Line q[N];Point p[N];
int h=0,t=0,k=0;q[t++]=l[0];
for(int i=1;i<n;i++)
{
if(sgn(l[i].angle()-l[i-1].angle())==0) continue;
while(h<t-1&&l[i].relation(q[t-1].crosspoint(q[t-2]))==2) t--;
while(h<t-1&&l[i].relation(q[h].crosspoint(q[h+1]))==2) h++;
q[t++]=l[i];
}
while(h<t-1&&l[h].relation(q[t-1].crosspoint(q[t-2]))==2) t--;
q[t++]=q[h];
for(int i=h;i<t-1;i++) p[k++]=q[i].crosspoint(q[i+1]);
//for(int i=0;i<k;i++) cout<<p[i].x<<" "<<p[i].y<<endl;
for(int i=1;i<k-1;i++) res+=(p[i]-p[0])^(p[i+1]-p[0]);
return res/2;
}
//`水平可见直线 从上向下看输出能看见哪些直线`
void half_plane2(Line l[],int n)
{
sort(l,l+n);
Line q[N];
int h=0,t=0,k=0;q[t++]=l[0];
for(int i=1;i<n;i++)
{
if(sgn(l[i].angle()-l[i-1].angle())==0) continue;
while(h<t-1&&l[i].relation(q[t-1].crosspoint(q[t-2]))==2) t--;
// while(h<t-1&&l[i].relation(q[h].crosspoint(q[h+1]))==2) h++;
q[t++]=l[i];
}
int ans[N];
for(int i=h;i<t;i++)
//for(auto j:q[i].id) ans[k++]=j;
sort(ans,ans+k);
cout<<k<<endl;
for(int i=0;i<k;i++) cout<<ans[i]<<" ";
}
//`多边形内核`
bool half_plane3(Line l[],int n)
{
sort(l,l+n);
Line q[N];
int h=0,t=0;q[t++]=l[0];
for(int i=1;i<n;i++)
{
if(sgn(l[i].angle()-l[i-1].angle())==0) continue;
while(h<t-1&&l[i].relation(q[t-1].crosspoint(q[t-2]))==2) t--;
while(h<t-1&&l[i].relation(q[h].crosspoint(q[h+1]))==2) h++;
q[t++]=l[i];
}
while(h<t-1&&l[h].relation(q[t-1].crosspoint(q[t-2]))==2) t--;
return t-h>=3;
}
//`最小圆覆盖`
circle increment(Point p[],int n)
{
random_shuffle(p,p+n);
circle ans;ans.p=p[0],ans.r=0;
for(int i=1;i<n;i++)
if(ans.r<ans.p.distance(p[i]))
{
ans.p=p[i],ans.r=0;
for(int j=0;j<i;j++)
if(ans.r<ans.p.distance(p[j]))
{
ans.p=(p[i]+p[j])/2,ans.r=p[i].distance(p[j])/2;
for(int k=0;k<j;k++)
if(ans.r<ans.p.distance(p[k]))
{
Point p1=(p[i]+p[j])/2;
Point v1=(p[i]-p[j]).rotright();
Point p2=(p[i]+p[k])/2;
Point v2=(p[i]-p[k]).rotright();
ans.p=Line(p1,p1+v1).crosspoint(Line(p2,p2+v2));
ans.r=ans.p.distance(p[i]);
}
}
}
return ans;
}
//`自适应辛普森积分`
inline double f(double x){ //积分函数
return x;
}
double simpson(double l,double r){//辛普森公式
double mid=(l+r)/2;
return (r-l)*(f(l)+f(r)+4*f(mid))/6;
}
double asr(double l,double r,double ans){//自适应
double mid=(l+r)/2;
double a=simpson(l,mid),b=simpson(mid,r);
if(sgn(a+b-ans)==0) return ans;
return asr(l,mid,a)+asr(mid,r,b);
}
void solve(int cnt)
{
double xl,yl,xr,yr;
cin>>xl>>yl>>xr>>yr;
Line l1,l2,l[N];
l1.input(),l2.input();
l1.adjust(),l2.adjust();
int t=l1.check(l2);
double res=0;
l[0]=Line(Point(xl,yl),Point(xr,yl));
l[1]=Line(Point(xr,yl),Point(xr,yr));
l[2]=Line(Point(xr,yr),Point(xl,yr));
l[3]=Line(Point(xl,yr),Point(xl,yl));
//cout<<t<<" ";
if(cnt==23) cout<<t<<" ";
if(t==1) res=(xr-xl)*(yr-yl);
else if(t==3)
{
if(sgn(l1.length()-l2.length())>0) swap(l1,l2);
if(l1.s==l2.s) l[4]=Line(l1.e,l2.e.rotate(l1.e,pi/2));
else if(l1.e==l2.e) l[4]=Line(l1.s,l2.s.rotate(l1.s,pi/2));
res=half_plane1(l,5);
}
else if(t==4)
{
if(l2.s<l1.s) swap(l1,l2);
if(l2.e<l1.e)
{
l[4]=Line(l2.s,l1.s.rotate(l2.s,pi/2));
l[5]=Line(l2.e,l1.e.rotate(l2.e,pi/2));
}
else
{
l[4]=Line(l2.s,l1.s.rotate(l2.s,pi/2));
l[5]=Line(l1.e,l2.e.rotate(l1.e,pi/2));
}
res=half_plane1(l,6);
}
else if(t==2)
{
if(l1.s==l2.s)
{
if(l1.e.y<l2.e.y) swap(l1,l2);
l[4]=Line(l1.s,l1.e.rotate(l1.s,pi/2));
l[5]=Line(l2.s,l2.e.rotate(l2.s,pi/2));
}
else if(l1.e==l2.e)
{
if(l1.s.y<l2.s.y) swap(l1,l2);
l[4]=Line(l1.e,l1.s.rotate(l1.e,pi/2));
l[5]=Line(l2.e,l2.s.rotate(l2.e,pi/2));
}
else if(l1.s==l2.e)
{
if(l1.e.y<l2.s.y) swap(l1,l2);
l[4]=Line(l1.e,l1.s.rotate(l1.e,pi/2));
l[5]=Line(l2.s,l2.e.rotate(l2.s,pi/2));
}
else
{
if(l1.s.y<l2.e.y) swap(l1,l2);
l[4]=Line(l1.e,l1.s.rotate(l1.e,pi/2));
l[5]=Line(l2.s,l2.e.rotate(l2.s,pi/2));
}
// cout<<l[4].s.x<<" "<<l[4].s.y<<endl;
// cout<<l[4].e.x<<" "<<l[4].e.y<<endl;
// cout<<l[5].s.x<<" "<<l[5].s.y<<endl;
// cout<<l[5].e.x<<" "<<l[5].e.y<<endl;
res=half_plane1(l,6);
}
cout<<fixed<<setprecision(10)<<res<<endl;
}
int main()
{
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int t=1; cin>>t;
for(int i=1;i<=t;i++) solve(i);
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 4048kb
input:
2 0 0 3 3 1 1 1 2 2 1 2 2 0 0 3 3 1 1 1 2 1 2 2 2
output:
0.0000000000 1.0000000000
result:
ok 2 numbers
Test #2:
score: 0
Accepted
time: 0ms
memory: 4356kb
input:
10 0 0 7 6 2 4 4 4 3 2 5 2 0 0 7 6 2 4 4 4 4 4 5 2 0 0 2 4 1 1 1 2 1 2 1 3 0 0 2 3 1 1 1 2 1 1 1 2 0 0 3 3 1 1 2 2 1 2 2 1 0 0 2 4 1 1 1 2 1 2 1 3 0 0 6 6 1 1 5 5 1 5 3 3 0 0 2 3 1 1 1 2 1 1 1 2 0 0 2 5 1 1 1 3 1 2 1 4 0 0 2 4 1 1 1 3 1 1 1 2
output:
0.0000000000 3.7500000000 0.0000000000 6.0000000000 0.0000000000 0.0000000000 0.0000000000 6.0000000000 2.0000000000 4.0000000000
result:
ok 10 numbers
Test #3:
score: -100
Wrong Answer
time: 245ms
memory: 4272kb
input:
100000 350 720 355 732 353 725 352 729 354 721 353 725 -807 606 -801 621 -803 608 -803 616 -803 616 -803 614 -389 463 -373 466 -382 464 -387 464 -387 464 -385 464 -664 801 -655 806 -656 803 -659 803 -659 803 -657 802 896 -767 901 -762 900 -763 897 -763 900 -763 897 -763 403 645 407 652 406 647 405 6...
output:
0.0000000000 42.0000000000 12.0000000000 24.0000000000 25.0000000000 28.0000000000 99.0000000000 0.0000000000 135.0000000000 6.0000000000 42.0000000000 45.0000000000 120.0000000000 8.0000000000 84.0000000000 15.0000000000 16.0000000000 0.0000000000 36.0000000000 4.0000000000 0.5000000000 20.00000000...
result:
wrong answer 23rd numbers differ - expected: '1.0000000', found: '2.0000000', error = '1.0000000'