题目大意:在二维平面中给出 n 个点,再给出一个固定大小的圆,问如何放置这个圆可以使其覆盖最多的点
题目分析:首先不难想到一种 n^3 的做法,就是两层循环去枚举两个点,因为两个不同的点就可以确定下来两个圆了(对称的),然后对于这 2 * n * n 个圆的每一个来说,再用一层循环去枚举 n 个点,计算一下有多少个点可以被覆盖到就可以了
考虑优化,假如分别以点 i 和点 j 为圆心,以 r 为半径做出两个相交的圆,比较显然的是,如果在相交的阴影部分中任选一点作为圆心,同样以 r 作为半径做圆,那么点 i 和点 j 都可以同时被覆盖到,如下图所示:
所以我们不妨映射到其中一个圆的弧上,称这一段为相交弧,这样一来此题就得以优化了:
先用一层循环去固定点 i 作为圆心,然后枚举点 j 同样也作为圆心,两个圆若能相交的话求出相交弧,最多有 n 段相交弧,对于以点 i 为圆心的圆周来说,其中某个点被覆盖的次数,就是以该点为圆心所能覆盖的点数,所以求出被覆盖最多的位置即可
如何去求这个位置呢?利用差分的思想,对 n 段相交弧,也就是 2 * n 个交点进行极角排序,然后扫一遍求最大连续子段和就是答案了,时间复杂度为 n^2logn
很让人烦心的一点是,这个题目用 atan2 的极角排序很轻松 AC,但用叉积排序总是多多少少会出现一些不可描述的问题,一直卡在 97 分的位置,纠结三天了,没精力再耗下去了。。就这样随缘吧
代码:
n^3
//#pragma GCC optimize(2)
//#pragma GCC optimize("Ofast","inline","-ffast-math")
//#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int N=110;
// `计算几何模板`
const double eps = 1e-8;
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;
}
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);
}
//`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);
}
}point[N];
//圆
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;
}
//输入
void input(){
p.input();
scanf("%lf",&r);
}
//输出
void output(){
printf("%.2lf %.2lf %.2lf\n",p.x,p.y,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));
}
//`点和圆的关系`
//`0 圆外`
//`1 圆上`
//`2 圆内`
int relation(Point b){
double dst = b.distance(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r)==0)return 1;
return 0;
}
//`两圆的关系`
//`5 相离`
//`4 外切`
//`3 相交`
//`2 内切`
//`1 内含`
//`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;
if(sgn(d-r-v.r) == 0)return 4;
double l = fabs(r-v.r);
if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;
if(sgn(d-l)==0)return 2;
if(sgn(d-l)<0)return 1;
}
//`求两个圆的交点,返回0表示没有交点,返回1是一个交点,2是两个交点`
//`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;
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;
return 2;
}
//`得到过a,b两点,半径为r1的两个圆`
static 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 = r1;
return t;
}
};
int main()
{
#ifndef ONLINE_JUDGE
// freopen("data.in.txt","r",stdin);
// freopen("data.out.txt","w",stdout);
#endif
// ios::sync_with_stdio(false);
double r;
scanf("%lf",&r);
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
point[i].input();
int ans=1;
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
{
circle c1,c2;
circle::gercircle(point[i],point[j],r,c1,c2);
int sum1=0,sum2=0;
for(int k=1;k<=n;k++)
{
if(c1.relation(point[k]))
sum1++;
if(c2.relation(point[k]))
sum2++;
}
ans=max(ans,sum1);
ans=max(ans,sum2);
}
printf("%d\n",ans);
return 0;
}
n^2logn
//#pragma GCC optimize(2)
//#pragma GCC optimize("Ofast","inline","-ffast-math")
//#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int N=110;
const double eps=1e-8;
int sgn(double x)
{
if(fabs(x)<=eps)
return 0;
if(x<0)
return -1;
return 1;
}
struct Point
{
double x,y;
void input()
{
scanf("%lf%lf",&x,&y);
}
double distance(const Point& t)const
{
return hypot(x-t.x,y-t.y);
}
}point[N];
struct Node
{
double alpha;
int flag;
Node(double alpha,int flag):alpha(alpha),flag(flag){}
bool friend operator<(const Node& a,const Node& b)
{
if(sgn(a.alpha-b.alpha)==0)
return a.flag>b.flag;
return sgn(a.alpha-b.alpha)<0;
}
};
int main()
{
#ifndef ONLINE_JUDGE
// freopen("data.in.txt","r",stdin);
// freopen("data.out.txt","w",stdout);
#endif
// ios::sync_with_stdio(false);
double r;
scanf("%lf",&r);
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
point[i].input();
int ans=1;
for(int i=1;i<=n;i++)
{
vector<Node>node;
for(int j=1;j<=n;j++)
{
if(i==j)
continue;
double dis=point[i].distance(point[j]);
if(sgn(dis-2*r)>0)
continue;
double alpha=atan2(point[j].y-point[i].y,point[j].x-point[i].x);
double phi=acos(dis/(2.0*r));
node.push_back(Node(alpha-phi,1));
node.push_back(Node(alpha+phi,-1));
}
sort(node.begin(),node.end());
int sum=0;
for(auto it:node)
{
sum+=it.flag;
ans=max(ans,sum+1);
}
}
printf("%d\n",ans);
return 0;
}