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POJ 3384 Feng Shui [半平面交]
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 5537 | Accepted: 1663 | Special Judge | ||
Description
Feng shui is the ancient Chinese practice of placement and arrangement of space to achieve harmony with the environment. George has recently got interested in it, and now wants to apply it to his home and bring harmony to it.
There is a practice which says that bare floor is bad for living area since spiritual energy drains through it, so George purchased two similar round-shaped carpets (feng shui says that straight lines and sharp corners must be avoided). Unfortunately, he is unable to cover the floor entirely since the room has shape of a convex polygon. But he still wants to minimize the uncovered area by selecting the best placing for his carpets, and asks you to help.
You need to place two carpets in the room so that the total area covered by both carpets is maximal possible. The carpets may overlap, but they may not be cut or folded (including cutting or folding along the floor border) — feng shui tells to avoid straight lines.
Input
The first line of the input file contains two integer numbers n and r — the number of corners in George’s room (3 ≤ n ≤ 100) and the radius of the carpets (1 ≤ r ≤ 1000, both carpets have the same radius). The following nlines contain two integers xi and yi each — coordinates of the i-th corner (−1000 ≤ xi, yi ≤ 1000). Coordinates of all corners are different, and adjacent walls of the room are not collinear. The corners are listed in clockwise order.
Output
Write four numbers x1, y1, x2, y2 to the output file, where (x1, y1) and (x2, y2) denote the spots where carpet centers should be placed. Coordinates must be precise up to 4 digits after the decimal point.
If there are multiple optimal placements available, return any of them. The input data guarantees that at least one solution exists.
题意：
在一个有 n 个点的凸包形状房间中，要放两个半径为 r 的圆形地毯，要求这两个地毯覆
盖地板的面积越大越好，求两个地毯圆心的位置。
先把凸包缩小r，就变成找圆心了，当然是最远点了，直接枚举点对就好了
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>using namespace std;typedef long long ll;const int N=205;const double INF=1e3+5;const double eps=1e-10;inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();} return x*f;}inline int sgn(double x){ if(abs(x)<eps) return 0; else return x<0?-1:1;}struct Vector{ double x,y; Vector(double a=0,double b=0):x(a),y(b){} bool operator <(const Vector &a)const{ return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0); } void print(){printf("%lf %lf\n",x,y);}};typedef Vector Point;Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);}Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;}double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}double Len(Vector a){return sqrt(Dot(a,a));}Vector Normal(Vector a){ return Vector(-a.y,a.x);//counterClockwise}struct Line{ Point s,t; Line(){} Line(Point a,Point b):s(a),t(b){}};bool isLSI(Line l1,Line l2){ Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s; return sgn(Cross(v,u))!=sgn(Cross(v,w));}Point LI(Line a,Line b){ Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s; double t=Cross(v2,v)/Cross(v1,v2); return a.s+v1*t;}void iniPolygon(Point p[],int &n,double inf){ n=0; p[++n]=Point(-inf,-inf); p[++n]=Point(inf,-inf); p[++n]=Point(inf,inf); p[++n]=Point(-inf,inf);}Point t[N];int tn;void CutPolygon(Point p[],int &n,Point a,Point b){//get the left of a->b tn=0; Point c,d; for(int i=1;i<=n;i++){ c=p[i],d=p[i%n+1]; if(sgn(Cross(b-a,c-a))>=0) t[++tn]=c; if(isLSI(Line(a,b),Line(c,d))) t[++tn]=LI(Line(a,b),Line(c,d)); } n=tn;for(int i=1;i<=n;i++) p[i]=t[i];}int n,r,m;Point p[N],q[N];Line L[N];void ChangePolygon(Point p[],int n,double x){ p[n+1]=p[1]; for(int i=1;i<=n;i++){ Vector v=Normal(p[i+1]-p[i])*x/Len(p[i+1]-p[i]); L[i]=Line(p[i]+v,p[i+1]+v); }}int main(int argc, const char * argv[]){ while(scanf("%d%d",&n,&r)!=EOF){ for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y); iniPolygon(q,m,INF); reverse(p+1,p+1+n); ChangePolygon(p,n,r); Point a,b;double mx=0; for(int i=1;i<=n;i++) CutPolygon(q,m,L[i].s,L[i].t); for(int i=1;i<=m;i++) for(int j=1;j<=m;j++) if(sgn(Len(q[j]-q[i])-mx)>=0){ mx=Len(q[j]-q[i]); a=q[i];b=q[j]; } if(a.x>b.x) swap(a,b); printf("%.4f %.4f %.4f %.4f\n",a.x,a.y,b.x,b.y); }}
POJ 3384 Feng Shui [半平面交]