poj 3384 Feng Shui(半平面交推进多边形)

题意：有一个多边形，要求你往里面放两个半径为r的圆，圆可以重合，要求的就是使得圆覆盖面积最大时的两个圆心坐标（SPJ输出任意一组就可以了）

思路：用半平面交推进r距离，得到的核区域就是圆心能够放置的区域，要使得圆覆盖面积最大，则就是圆心距离最远，知道了这个然后求一组最远点对，求出来的就是一组圆心的可行解

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cstring>#include<cmath>#include<algorithm>#define inf 100000000using namespace std;const double pi=acos(-1.0);const double eps=1e-8;int n,ln,dq[100000],top,bot,pn;struct P{double x,y;P (){}P(double x,double y):x(x),y(y){}P operator +(P p){return P(x+p.x,y+p.y);}P operator -(P p){return P(x-p.x,y-p.y);}P operator *(double d){return P(x*d,y*d);}double det(P p){return x*p.y-y*p.x;}};P p[10000000];struct Line{P a,b;double angle;};Line l[100000],l1[100000];P in(Line a,Line b){return a.a+(a.b-a.a)*((b.b-b.a).det(b.a-a.a) / (b.b-b.a).det(a.b-a.a));}int dblcmp(double k){if(fabs(k)<eps)return 0;return k>0?1:-1;}bool cmp(Line a,Line b){int d=dblcmp(a.angle-b.angle);if(!d)return dblcmp((b.a-a.a).det(b.b-a.a))<0;return d<0;}bool judge(Line a,Line b,Line c){P r=in(b,c);return dblcmp((a.a-r).det(a.b-r))>0; //顺时针大于0，逆时针小于0}double dis(P a,P b){double c=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);return sqrt(c);}//向量向内侧平移r距离函数(顺逆时针时函数不一样，此处为顺时针，逆时针改动一小部分就可以了)void getmove(double r){for(int i=0;i<ln;i++){double x=l[i].a.x-l[i].b.x;double y=l[i].a.y-l[i].b.y;double L=dis(l[i].a,l[i].b);l1[i].a.x=(l[i].a.x-r*y/L);l1[i].a.y=(l[i].a.y+r*x/L);l1[i].b.x=(l[i].b.x-r*y/L);l1[i].b.y=(l[i].b.y+r*x/L);l1[i].angle=l[i].angle;}}P r1,r2;void check(double r){int len;int i,j;getmove(r);for(i=0,j=0;i<ln;i++){if(dblcmp(l1[i].angle-l1[j].angle)>0)l1[++j]=l1[i];}len=j+1;top=1;bot=0;dq[0]=0;dq[1]=1;for(i=2;i<len;i++){while(top>bot && judge(l1[i],l1[dq[top]],l1[dq[top-1]])) top--;while(top>bot && judge(l1[i],l1[dq[bot]],l1[dq[bot+1]])) bot++;dq[++top]=i;}while(top>bot && judge(l1[dq[bot]],l1[dq[top]],l1[dq[top-1]])) top--;while(top>bot && judge(l1[dq[top]],l1[dq[bot]],l1[dq[bot+1]])) bot++;dq[++top]=dq[bot];double mxn=-10000;for(i=bot,pn=0;i<top;i++,pn++){p[pn]=in(l1[dq[i]],l1[dq[i+1]]);//printf("%lf %lf\n",p[pn].x,p[pn].y);}//因为有可能核为一个点，所以直接枚举的所有交点for(i=0;i<pn;i++){for(int j=i+1;j<pn;j++){if(dis(p[i],p[j])>mxn){mxn=dis(p[i],p[j]);r1=p[i];r2=p[j];}}}}int main(){double r;while(~scanf("%d%lf",&n,&r) && n){ln=0;for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);for(int i=0;i<n-1;i++){l[ln].a=p[i];l[ln].b=p[i+1];l[ln].angle=atan2(l[ln].b.y-l[ln].a.y,l[ln].b.x-l[ln].a.x);ln++;}l[ln].a=p[n-1];l[ln].b=p[0];l[ln].angle=atan2(l[ln].b.y-l[ln].a.y,l[ln].b.x-l[ln].a.x);ln++;sort(l,l+ln,cmp);check(r);printf("%.4lf %.4lf %.4lf %.4lf\n",r1.x,r1.y,r2.x,r2.y);}return 0;}