题目给出两个圆和一个多边形
问是否能让两个圆在多边形内。
并且覆盖的面积最大
圆的半径为r，我们则让多边形的每条边都往内部退r距离。
然后求半平面交得出的点集中，最远的两个点则是两圆的圆心即可
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <sstream>
#include <queue>
#include <vector>
#define MAXN 111111
#define MAXM 211111
#define PI acos(-1.0)
#define eps 1e-8
#define INF 1000000001
using namespace std;
int dblcmp(double d)
{
if (fabs(d) < eps) return 0;
return d > eps ? 1 : -1;
}
struct point
{
double x, y;
point(){}
point(double _x, double _y):
x(_x), y(_y){};
void input()
{
scanf("%lf%lf",&x, &y);
}
double dot(point p)
{
return x * p.x + y * p.y;
}
double distance(point p)
{
return hypot(x - p.x, y - p.y);
}
point sub(point p)
{
return point(x - p.x, y - p.y);
}
double det(point p)
{
return x * p.y - y * p.x;
}
bool operator == (point a)const
{
return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
}
bool operator < (point a)const
{
return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
}
}p[MAXN];
struct line
{
point a,b;
line(){}
line(point _a,point _b)
{
a=_a;
b=_b;
}
bool parallel(line v)
{
return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0;
}
point crosspoint(line v)
{
double a1 = v.b.sub(v.a).det(a.sub(v.a));
double a2 = v.b.sub(v.a).det(b.sub(v.a));
return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));
}
bool operator == (line v)const
{
return (a == v.a) && (b == v.b);
}
};
struct halfplane:public line
{
double angle;
halfplane(){}
//表示向量 a->b逆时针(左侧)的半平面
halfplane(point _a, point _b)
{
a = _a;
b = _b;
}
halfplane(line v)
{
a = v.a;
b = v.b;
}
void calcangle()
{
angle = atan2(b.y - a.y, b.x - a.x);
}
bool operator <(const halfplane &b)const
{
return angle < b.angle;
}
};
struct polygon
{
int n;
point p[MAXN];
line l[MAXN];
double area;
void getline()
{
for (int i = 0; i < n; i++)
{
l[i] = line(p[i], p[(i + 1) % n]);
}
}
void getarea()
{
area = 0;
int a = 1, b = 2;
while(b <= n - 1)
{
area += p[a].sub(p[0]).det(p[b].sub(p[0]));
a++;
b++;
}
area = fabs(area) / 2;
}
}convex;
struct halfplanes
{
int n;
halfplane hp[MAXN];
point p[MAXN];
int que[MAXN];
int st, ed;
void push(halfplane tmp)
{
hp[n++] = tmp;
}
void unique()
{
int m = 1, i;
for (i = 1; i < n;i++)
{
if (dblcmp(hp[i].angle - hp[i - 1].angle))hp[m++] = hp[i];
else if (dblcmp(hp[m - 1].b.sub(hp[m - 1].a).det(hp[i].a.sub(hp[m - 1].a)) > 0))hp[m - 1] = hp[i];
}
n = m;
}
bool halfplaneinsert()
{
int i;
for (i = 0; i < n; i++) hp[i].calcangle();
sort(hp, hp + n);
unique();
que[st = 0] = 0;
que[ed = 1] = 1;
p[1] = hp[0].crosspoint(hp[1]);
for (i = 2; i < n; i++)
{
while (st < ed && dblcmp((hp[i].b.sub(hp[i].a).det(p[ed].sub(hp[i].a)))) < 0) ed--;
while (st < ed && dblcmp((hp[i].b.sub(hp[i].a).det(p[st + 1].sub(hp[i].a)))) < 0) st++;
que[++ed] = i;
if (hp[i].parallel(hp[que[ed - 1]])) return false;
p[ed] = hp[i].crosspoint(hp[que[ed - 1]]);
}
while (st < ed && dblcmp(hp[que[st]].b.sub(hp[que[st]].a).det(p[ed].sub(hp[que[st]].a))) < 0) ed--;
while (st < ed && dblcmp(hp[que[ed]].b.sub(hp[que[ed]].a).det(p[st + 1].sub(hp[que[ed]].a))) < 0) st++;
if (st + 1 >= ed)return false;
return true;
}
void getconvex(polygon &con)
{
p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
con.n = ed - st + 1;
int j = st, i = 0;
for (; j <= ed; i++, j++)
{
con.p[i] = p[j];
}
}
}h;
int T;
int n;
line getmove(point a, point b, double mid)
{
double x = a.x - b.x;
double y = a.y - b.y;
double L = a.distance(b);
point ta = point(mid * y / L + a.x, a.y - mid * x / L);
point tb = point(mid * y / L + b.x, b.y - mid * x / L);
return line(ta, tb);
}
double r;
int main()
{
int cas = 0;
while(scanf("%d%lf", &n, &r) != EOF)
{
for(int i = 0; i < n; i++) p[i].input();
h.n = 0;
for(int i = 0; i < n; i++)
{
line tmp = getmove(p[(i + 1) % n], p[i], r);
h.push(halfplane(tmp));
}
h.halfplaneinsert();
h.getconvex(convex);
int id1 = 0, id2 = 0;
double mx = 0;
for(int i = 0; i < convex.n; i++)
for(int j = i + 1; j < convex.n; j++)
{
double len = convex.p[i].distance(convex.p[j]);
if(dblcmp(len - mx) > 0) id1 = i, id2 = j, mx = len;
}
printf("%f %f %f %f
", convex.p[id1].x, convex.p[id1].y, convex.p[id2].x, convex.p[id2].y);
}
return 0;
}