之前gym做过一个画圆的。。。
然后这个也想画圆。。。
算了算复杂度n^3logn还蛮好。
卡常卡了3个小时投降了。
学艺不精啊。。。
赛后去看了qls的代码。学到了神奇的优化技巧。
比如我们要check点到直线的距离是否大于等于二分的半径r。
我们可以直接用叉积判，而不是算投影。。。
我觉得最大的优化就在这了，理论上省了一半多常数。
之前给学弟讲课还说了计算几何最重要的就是会用叉积结果自己都不会。。
题解的结论其实不是很难想，奈何先入为主地想到了其他理论上很对的做法也就不会去猜什么结论了。。
哭了。300的数据n^3logn还要精打细算的卡常。。。
扔个画圆的代码。
#include <bits/stdc++.h>
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
void prt(db x){printf("%11f
",x);}
inline int sign(db k){
if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
inline 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;}
inline 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);}
inline 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;}
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;}
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]);}
int parallel(line k1,line k2){return sign(cross(k1.dir(),k2.dir()))==0;}
int sameDir(line k1,line k2){return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==1;}
int operator < (line k1,line k2){
if (sameDir(k1,k2)) return k2.include(k1[0]);
return compareangle(k1.dir(),k2.dir());
}
int checkpos(line k1,line k2,line k3){return k3.include(getLL(k1,k2));}
vector<line> getHL(vector<line> &L){ // 求半平面交 , 半平面是逆时针方向 , 输出按照逆时针
sort(L.begin(),L.end()); deque<line> q;
for (int i=0;i<(int)L.size();i++){
if (i&&sameDir(L[i],L[i-1])) continue;
while (q.size()>1&&!checkpos(q[q.size()-2],q[q.size()-1],L[i])) q.pop_back();
while (q.size()>1&&!checkpos(q[1],q[0],L[i])) q.pop_front();
q.push_back(L[i]);
}
while (q.size()>2&&!checkpos(q[q.size()-2],q[q.size()-1],q[0])) q.pop_back();
while (q.size()>2&&!checkpos(q[1],q[0],q[q.size()-1])) q.pop_front();
vector<line>ans; for (int i=0;i<q.size();i++) ans.push_back(q[i]);
return ans;
}
db closepoint(vector<point>&A,int l,int r){ // 最近点对 , 先要按照 x 坐标排序
if (r-l<=5){
db ans=1e20;
for (int i=l;i<=r;i++) for (int j=i+1;j<=r;j++) ans=min(ans,A[i].dis(A[j]));
return ans;
}
int mid=l+r>>1; db ans=min(closepoint(A,l,mid),closepoint(A,mid+1,r));
vector<point>B; for (int i=l;i<=r;i++) if (abs(A[i].x-A[mid].x)<=ans) B.push_back(A[i]);
sort(B.begin(),B.end(),[](point k1,point k2){return k1.y<k2.y;});
for (int i=0;i<B.size();i++) for (int j=i+1;j<B.size()&&B[j].y-B[i].y<ans;j++) ans=min(ans,B[i].dis(B[j]));
return ans;
}
int checkposCC(circle k1,circle k2){// 返回两个圆的公切线数量
if (cmp(k1.r,k2.r)==-1) swap(k1,k2);
db dis=k1.o.dis(k2.o); int w1=cmp(dis,k1.r+k2.r),w2=cmp(dis,k1.r-k2.r);
if (w1>0) return 4; else if (w1==0) return 3; else if (w2>0) return 2;
else if (w2==0) return 1; else return 0;
}
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};
}
vector<point> getCC(circle k1,circle k2){// 沿圆 k1 逆时针给出 , 相切给出两个
int pd=checkposCC(k1,k2); if (pd==0||pd==4) return {};
db a=(k2.o-k1.o).abs2(),cosA=(k1.r*k1.r+a-k2.r*k2.r)/(2*k1.r*sqrt(max(a,(db)0.0)));
db b=k1.r*cosA,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
point k=(k2.o-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
return {m-del,m+del};
}
vector<point> TangentCP(circle k1,point k2){// 沿圆 k1 逆时针给出
db a=(k2-k1.o).abs(),b=k1.r*k1.r/a,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
point k=(k2-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
return {m-del,m+del};
}
inline vector<line> TangentoutCC(circle k1,circle k2){
int pd=checkposCC(k1,k2); if (pd==0) return {};
if (pd==1){point k=getCC(k1,k2)[0]; return {(line){k,k}};}
if (cmp(k1.r,k2.r)==0){
point del=(k2.o-k1.o).unit().turn90().getdel();
return {(line){k1.o-del*k1.r,k2.o-del*k2.r},(line){k1.o+del*k1.r,k2.o+del*k2.r}};
} else {
point p=(k2.o*k1.r-k1.o*k2.r)/(k1.r-k2.r);
vector<point>A=TangentCP(k1,p),B=TangentCP(k2,p);
vector<line>ans; for (int i=0;i<A.size();i++) ans.push_back((line){A[i],B[i]});
return ans;
}
}
inline vector<line> TangentinCC(circle k1,circle k2){
int pd=checkposCC(k1,k2); if (pd<=2) return {};
if (pd==3){point k=getCC(k1,k2)[0]; return {(line){k,k}};}
point p=(k2.o*k1.r+k1.o*k2.r)/(k1.r+k2.r);
vector<point>A=TangentCP(k1,p),B=TangentCP(k2,p);
vector<line>ans; for (int i=0;i<A.size();i++) ans.push_back((line){A[i],B[i]});
return ans;
}
inline vector<line> TangentCC(circle k1,circle k2){
int flag=0; if (k1.r<k2.r) swap(k1,k2),flag=1;
vector<line>A=TangentoutCC(k1,k2),B=TangentinCC(k1,k2);
for (line k:B) A.push_back(k);
if (flag) for (line &k:A) swap(k[0],k[1]);
return A;
}
int n;
circle c[305];
vector<line> v;
bool check(db d){
for(int i=1;i<=n;i++)c[i].r=d;
for(int i=1;i<n;i++){
for(int j=i+1;j<=n;j++){
v=TangentinCC(c[i],c[j]);
for(auto l:v){
bool f = 1;
int x=0,y=0;
point dir = l[1] - l[0];
db s = dir.abs() * d;
for (int k=1;k<=n&&x<=n/2-1&&y<=n/2-1&&f;k++){
if(k==i||k==j)continue;
db nmsl = cross(dir, c[k].o - l[0]);
f &= (cmp(fabs(nmsl),s)>=0);
if(nmsl>0)x++;
else y++;
}
if(f&&x==y)return true;
}
}
}
return false;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lf%lf",&c[i].o.x,&c[i].o.y);
}
db l=0,r=2e6;
for(int i=1;i<=42;i++){
db mid = (l+r)/2;
if(check(mid)){
l=mid;
}else{
r=mid;
}
}
printf("%.11f
",l);
}
扔个题解思路代码
#include <bits/stdc++.h>
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
inline int sign(db k){
if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
inline int cmp(db k1,db k2){return sign(k1-k2);}
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};}
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 cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
int n;point p[1005];
db ans = 0;db dis[1005];
void slove(point x){
for(int i=1;i<=n;i++){
dis[i]=cross(x,p[i])/x.abs();
}
sort(dis+1,dis+1+n);
ans = max(ans,fabs(dis[n/2+1]-dis[n/2])/2);
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lf%lf",&p[i].x,&p[i].y);
}
for(int i=1;i<n;i++){
for(int j=i+1;j<=n;j++){
point dir=p[j]-p[i];
slove(dir);
slove(dir.turn90());
}
}
printf("%.12f
",ans);
}