HDU - 3644：A Chocolate Manufacturer's Problem（模拟退火， 求多边形内最大圆半径）

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=110;
const double pi=acos(-1.0);
const double inf=0x7fffffff;
struct point {
    double x,y;
    point(){}
    point(double xx,double yy):x(xx),y(yy){}
};
struct line{
    point s,p;
    line(){}
    line(point xx,point yy):s(xx),p(yy){}
};
double getdis(point w,point v){
    return sqrt((w.x-v.x)*(w.x-v.x)+(w.y-v.y)*(w.y-v.y));
}
point operator /(point a,double t){ return point(a.x/t,a.y/t);}
point operator *(point a,double t){ return point(t*a.x,t*a.y);}
point operator -(point w,point v){return point(w.x-v.x,w.y-v.y);}
point operator +(point w,point v){return point(w.x+v.x,w.y+v.y);}
double det(point w,point v){ return w.x*v.y-w.y*v.x;}
double dot(point w,point v){ return w.x*v.x+w.y*v.y;}
double ltoseg(point p,point a,point b){
    point t=p-a;
    if(dot(t,b-a)<=0) return getdis(p,a);
    else if(dot(p-b,a-b)<=0) return getdis(p,b);
    return fabs(det(t,b-a))/getdis(a,b);
}
point p[maxn],tp[maxn]; double dist[maxn]; int N; line L[maxn];
bool isinside(point a)
{
//算法描述：首先，对于多边形的水平边不做考虑，其次，
//对于多边形的顶点和射线相交的情况，如果该顶点时其所属的边上纵坐标较大的顶点，则计数，否则忽略该点，
//最后，对于Q在多边形上的情形，直接判断Q是否属于多边形。
    int ncross=0;
    rep(i,0,N-1) {
        point p1=p[i],p2=p[i+1];
        if(ltoseg(a,p[i],p[i+1])==0) return true; //在线段上
        if(p1.y==p2.y) continue; //默认做水平x轴的线，所以水平线不考虑
        if(a.y<min(p1.y,p2.y)) continue; //相离不考虑
        if(a.y>max(p1.y,p2.y)) continue;
        double t=det(a-p[i],a-p[i+1]);
        if((t>=0&&p[i].y<a.y&&p[i+1].y>=a.y)||(t<=0&&p[i+1].y<a.y&&p[i].y>=a.y)) ncross++;
    }
    return (ncross&1);
}
double getmindis(point a)
{
    double ans=inf;
    rep(i,0,N-1) ans=min(ans,ltoseg(a,p[i],p[i+1]));
    return ans;
}
int main()
{
    srand(unsigned(time(NULL)));
    while(~scanf("%d",&N)&&N){
        double X,Y,R; X=Y=0;
        rep(i,1,N) {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            X=max(X,p[i].x);
            Y=max(Y,p[i].y);
        }
        p[0]=p[N];
        rep(i,0,N-1) L[i]=line(p[i],p[i+1]-p[i]);
        scanf("%lf",&R);
        int maxt=min(N,20);
        rep(i,0,maxt-1){
            tp[i]=(p[i]+p[i+1])/2;
            dist[i]=0;
        }
        double step=min(X,Y);
        const int maxd=10;
        const double rate=0.55;
        bool flag=0;
        const double EPS2=1e-6;
        while(step>EPS2&&!flag){
            rep(i,0,maxt-1){
                rep(j,0,maxd-1){
                    double d=rand()%360/360.0*2*pi;
                    point next=tp[i];
                    next.x+=step*sin(d);
                    next.y+=step*cos(d);
                    if(!isinside(next)) continue;
                    double tdis=getmindis(next);
                    if(tdis+EPS2>dist[i]){
                        dist[i]=tdis; tp[i]=next;
                    }
                    if(tdis+EPS2>=R){
                        flag=1; break;
                    }
                }
            }
            step*=rate;
        }
        if(flag) puts("Yes");
        else puts("No");
    }
    return 0;
}