POJ 3384 Feng Shui(半平面交)

题意：在一个给定多边形里放两个半径为 r 的圆，问两个圆覆盖总面积最大时圆心坐标。

思路：把多边形每条边向内移动 r ，再做半平面交，得到的多边形就是可以放半径为 r 的圆的区域。

容易发现满足题意的两个圆一定是在顶点上，于是枚举即可。（这题是 Special Judge）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int maxpt = 220;
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;
}
Point operator -(const Point &b)const
{
return Point(x - b.x, y - b.y);
}
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
};
struct Line
{
Point s,e;
double k;
Line(){}
Line(Point _s,Point _e)
{
s = _s; e = _e;
k = atan2(e.y - s.y,e.x - s.x);
}
Point operator &(const Line &b)const
{
Point res = s;
double t = ((s - b.s)^(b.s - b.e))/((s - e)^(b.s - b.e));
res.x += (e.x - s.x)*t;
res.y += (e.y - s.y)*t;
return res;
}
};
//半平面交，直线的左边代表有效区域
bool HPIcmp(Line a,Line b)
{
if(fabs(a.k - b.k) > eps)return a.k < b.k;
return ((a.s - b.s)^(b.e - b.s)) < 0;
}
Line Q[maxpt];
void HPI(Line line[], int n, Point res[], int &resn)
{
int tot = n;
sort(line,line+n,HPIcmp);
tot = 1;
for(int i = 1;i < n;i++)
if(fabs(line[i].k - line[i-1].k) > eps)
line[tot++] = line[i];
int head = 0, tail = 1;
Q[0] = line[0];
Q[1] = line[1];
resn = 0;
for(int i = 2; i < tot; i++)
{
if(fabs((Q[tail].e-Q[tail].s)^(Q[tail-1].e-Q[tail-1].s)) < eps || fabs((Q[head].e-Q[head].s)^(Q[head+1].e-Q[head+1].s)) < eps)
return;
while(head < tail && (((Q[tail]&Q[tail-1]) - line[i].s)^(line[i].e-line[i].s)) > eps)
tail--;
while(head < tail && (((Q[head]&Q[head+1]) - line[i].s)^(line[i].e-line[i].s)) > eps)
head++;
Q[++tail] = line[i];
}
while(head < tail && (((Q[tail]&Q[tail-1]) - Q[head].s)^(Q[head].e-Q[head].s)) > eps)
tail--;
while(head < tail && (((Q[head]&Q[head-1]) - Q[tail].s)^(Q[tail].e-Q[tail].e)) > eps)
head++;
if(tail <= head + 1)return;
for(int i = head; i < tail; i++)
res[resn++] = Q[i]&Q[i+1];
if(head < tail - 1)
res[resn++] = Q[head]&Q[tail];
}
//*两点间距离
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
void change(Point a,Point b,Point &c,Point &d,double p)//将线段ab往左移动距离p
{
double len = dist(a,b);
double dx = (a.y - b.y)*p/len;
double dy = (b.x - a.x)*p/len;
c.x = a.x + dx; c.y = a.y + dy;
d.x = b.x + dx; d.y = b.y + dy;
}

Point pt[maxpt];
Line line[maxpt];
Point poly[maxpt];

int main()
{
int n;
double r;

while (~scanf("%d%lf", &n, &r)) {
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &pt[i].x, &pt[i].y);
if (i)
line[i] = Line(pt[i], pt[i-1]);
}
line[0] = Line(pt[0], pt[n-1]);
for (int i = 0; i < n; i++) {
Point p1 = line[i].s, p2 = line[i].e;
change(p1, p2, line[i].s, line[i].e, r);
}
int num;
Point rs1, rs2;
double cur = -1;
HPI(line, n, poly, num);
for (int i = 0; i < num; i++) {
for (int j = i + 1; j < num; j++) {
if (sgn(dist(poly[i], poly[j]) - cur) > 0) {
cur = dist(poly[i], poly[j]);
rs1 = poly[i], rs2 = poly[j];
}
}
}
printf("%lf %lf %lf %lf\n", rs1.x, rs1.y, rs2.x, rs2.y);
}

return 0;
}