发现自己对pslg这边不怎么会,打算对着蓝书刷几道康康
不过好像近几年确实很冷门虽然牛客多校出了个板子
这个就是简单的cut多边形然后暴力判就行了
这macOS 10.15到底有啥用啊,测试版就有clion不能debug的问题了,正式版竟然还有。。。
去jb看他们好像说在clion 9月17那个版本里把这个问题解决了?但是我下了一遍没用?
唔雨又下大了((
唔ummmm>_<
#include <bits/stdc++.h>
#define pii pair<int,int>
#define yxn inline
#define INF 19970404
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k){
if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内
struct point{
db x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
point operator * (db k1) const{return (point){x*k1,y*k1};}
point operator / (db k1) const{return (point){x/k1,y/k1};}
int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
// 逆时针旋转
point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
point turn90(){return (point){-y,x};}
bool operator < (const point k1) const{
int a=cmp(x,k1.x);
if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
}
db abs(){return sqrt(x*x+y*y);}
db abs2(){return x*x+y*y;}
db dis(point k1){return ((*this)-k1).abs();}
point unit(){db w=abs(); return (point){x/w,y/w};}
void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;}
void print(){printf("%.11lf %.11lf
",x,y);}
db getw(){return atan2(y,x);}
point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}
// -pi -> pi
int compareangle (point k1,point k2){//极角排序+
return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0);
}
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影
point k=k2-k1; return k1+k*(dot(q-k1,k)/k.abs2());
}
point reflect(point k1,point k2,point q){return proj(k1,k2,q)*2-q;}
int clockwise(point k1,point k2,point k3){// k1 k2 k3 逆时针 1 顺时针 -1 否则 0
return sign(cross(k2-k1,k3-k1));
}
int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点
return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=0;
}
point getLL(point k1,point k2,point k3,point k4){
db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
int intersect(db l1,db r1,db l2,db r2){
if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;
}
int checkSS(point k1,point k2,point k3,point k4){
return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}
db disSP(point k1,point k2,point q){
point k3=proj(k1,k2,q);
if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));
}
db disSS(point k1,point k2,point k3,point k4){
if (checkSS(k1,k2,k3,k4)) return 0;
else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));
}
int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}
db Area(vector<point> A){ // 多边形用 vector<point> 表示 , 逆时针
db ans=0;
for (int i=0;i<A.size();i++) ans+=cross(A[i],A[(i+1)%A.size()]);
return ans/2;
}
struct circle{
point o; db r;
void scan(){o.scan(); scanf("%lf",&r);}
int inside(point k){return cmp(r,o.dis(k));}
};
struct line{
// p[0]->p[1]
point p[2];
line(point k1,point k2){p[0]=k1; p[1]=k2;}
line(){}
point& operator [] (int k){return p[k];}
int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>0;}
point dir(){return p[1]-p[0];}
line push(){ // 向外 ( 左手边 ) 平移 eps
const db eps = 1e-6;
point delta=(p[1]-p[0]).turn90().unit()*eps;
return {p[0]-delta,p[1]-delta};
}
};
point getLL(line k1,line k2){return getLL(k1[0],k1[1],k2[0],k2[1]);}
vector<point> getCL(circle k1,point k2,point k3){ // 沿着 k2->k3 方向给出 , 相切给出两个
point k=proj(k2,k3,k1.o); db d=k1.r*k1.r-(k-k1.o).abs2();
if (sign(d)==-1) return {};
point del=(k3-k2).unit()*sqrt(max((db)0.0,d)); return {k-del,k+del};
}
int contain(vector<point>A,point q){ // 2 内部 1 边界 0 外部
int pd=0; A.push_back(A[0]);
for (int i=1;i<A.size();i++){
point u=A[i-1],v=A[i];
if (onS(u,v,q)) return 1; if (cmp(u.y,v.y)>0) swap(u,v);
if (cmp(u.y,q.y)>=0||cmp(v.y,q.y)<0) continue;
if (sign(cross(u-v,q-v))<0) pd^=1;
}
return pd<<1;
}
vector<point> convexcut(vector<point>A,point k1,point k2){//多边形
// 保留 k1,k2,p 逆时针的所有点
int n=A.size(); A.push_back(A[0]); vector<point>ans;
for (int i=0;i<n;i++){
int w1=clockwise(k1,k2,A[i]),w2=clockwise(k1,k2,A[i+1]);
if (w1>=0) ans.push_back(A[i]);
if (w1*w2<0) ans.push_back(getLL(k1,k2,A[i],A[i+1]));
}
return ans;
}
//---------板子分界线--------------------------------
int n,m,L,W;
circle c;
vector<line>l;
vector<vector<point>>ply,nply;
vector<point> v;
bool check(vector<point> v){
if(c.inside(v[0])==1)return true;
if(contain(v,c.o)>=1)return true;
//if(c.inside(v[0]))return true;
for(int i=0;i<v.size();i++){
vector<point> x = getCL(c,v[i],v[(i+1)%v.size()]);//
// for(auto p:x)if(onS(v[i],v[(i+1)%n],p)&&!(p==v[i])&&!(p==v[(i+1)%n]))return true;
if(cmp(disSP(v[i],v[(i+1)%v.size()],c.o),c.r)==-1){
return true;
}
}
return false;
}
void slove(){
vector<db> ans;
for(int i=0;i<ply.size();i++){
if(check(ply[i])){
//printf("%d
",i);
ans.push_back(abs(Area(ply[i])));
}
}
sort(ans.begin(),ans.end());
printf("%d",(int)ans.size());
for(auto x:ans){
printf(" %.2f",x);
}
printf("
");
}
void cut(line l){
nply.clear();
for(int i=0;i<ply.size();i++){
vector<point> x = convexcut(ply[i],l[0],l[1]);
vector<point> y = convexcut(ply[i],l[1],l[0]);
if(x.size()>=3)nply.push_back(x);
if(y.size()>=3)nply.push_back(y);
}
ply=nply;
}
int main(){
while(scanf("%d%d%d%d", &n, &m, &L, &W) == 4 && n) {
v.clear();ply.clear();nply.clear();
v.push_back({0,0});
v.push_back({1.0*L,0});
v.push_back({1.0*L,1.0*W});
v.push_back({0,1.0*W});
ply.push_back(v);
l.resize(n);
for(int i=0;i<n;i++){
scanf("%lf%lf%lf%lf",&l[i][0].x,&l[i][0].y,&l[i][1].x,&l[i][1].y);
cut(l[i]);
// printf("%d
",(int)ply.size());
}
for(int i=0;i<m;i++){
scanf("%lf%lf%lf",&c.o.x,&c.o.y,&c.r);
slove();
}
printf("
");
}
}