• POJ 3525 Most Distant Point from the Sea


    http://poj.org/problem?id=3525

    给出一个凸包,要求凸包内距离所有边的长度的最小值最大的是哪个

    思路:二分答案,然后把凸包上的边移动这个距离,做半平面交看是否有解。

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    const double finf=1e10;
    const double eps=1e-7;
    const double Pi=acos(-1);
    int n,tot;
    struct Point{
        double x,y;
        Point(){}
        Point(double x0,double y0):x(x0),y(y0){}
    }p[200005];
    struct Line{
        Point s,e;
        double slop;
        Line(){}
        Line(Point s0,Point e0):s(s0),e(e0){}
    }l[200005],L[200005],c[200005];
    int read(){
        int t=0,f=1;char ch=getchar();
        while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
        while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
        return t*f;
    }
    Point operator *(Point p,double x){
        return Point(p.x*x,p.y*x);
    }
    Point operator /(Point p,double x){
        return Point(p.x/x,p.y/x);
    }
    double operator *(Point p1,Point p2){
        return p1.x*p2.y-p1.y*p2.x;
    }
    Point operator -(Point p1,Point p2){
        return Point(p1.x-p2.x,p1.y-p2.y);
    }
    Point operator +(Point p1,Point p2){
        return Point(p1.x+p2.x,p1.y+p2.y);
    }
    double sqr(double x){
        return x*x;
    }
    double dis(Point p){
        return sqrt(sqr(p.x)+sqr(p.y));
    }
    Point e(Point p){
        double len=dis(p);
        p=p/len;
        return p;
    }
    Point turn(Point p,double x){
        double Sin=sin(x),Cos=cos(x);
        double X=Cos*p.x-Sin*p.y;
        double Y=Cos*p.y+Sin*p.x;
        return Point(X,Y);
    }
    bool cmp(Line p1,Line p2){
        if (p1.slop!=p2.slop) return p1.slop<p2.slop;
        else return (p1.e-p1.s)*(p2.e-p1.s)<=0;
    }
    void build(double mid){
        for (int i=1;i<=tot;i++){
         Point p=e(turn(l[i].e-l[i].s,Pi/2.0))*mid;
         L[i].s=l[i].s+p;
         L[i].e=l[i].e+p;
        }
        for (int i=1;i<=tot;i++)
         L[i].slop=l[i].slop;
        std::sort(L+1,L+1+tot,cmp);  
    }
    Point inter(Line p1,Line p2){
        double k1=(p2.e-p1.s)*(p1.e-p1.s);
        double k2=(p1.e-p1.s)*(p2.s-p1.s);
        double t=(k2/(k1+k2));
        double x=p2.s.x+(p2.e.x-p2.s.x)*t;
        double y=p2.s.y+(p2.e.y-p2.s.y)*t;
        return Point(x,y);
    }
    bool jud(Line p1,Line p2,Line p3){
        Point p=inter(p1,p2);
        return (p-p3.s)*(p3.e-p3.s)>0;
    }
    bool phi(){
        int cnt=1;
        for (int i=2;i<=tot;i++)
         if (L[i].slop!=L[i-1].slop) L[++cnt]=L[i];
        int lll=1,rrr=2;c[lll]=L[1];c[rrr]=L[2];
        for (int i=3;i<=cnt;i++){
            while (lll<rrr&&jud(c[rrr],c[rrr-1],L[i])) rrr--;
            while (lll<rrr&&jud(c[lll],c[lll+1],L[i])) lll++;
            c[++rrr]=L[i];
        } 
        while (lll<rrr&&jud(c[rrr],c[rrr-1],c[lll])) rrr--;
        while (lll<rrr&&jud(c[lll],c[lll+1],c[rrr])) lll++;
        if (rrr-lll+1>=3) return 1;
        else return 0;
    }
    bool check(double mid){
        build(mid);
        if (phi()) return 1;
        return 0;
    }
    int main(){
        while (scanf("%d",&n)!=EOF){
            if (n==0) return 0;
            for (int i=1;i<=n;i++)
             p[i].x=read(),p[i].y=read();
            p[n+1]=p[1];
            tot=0;
            for (int i=1;i<=n;i++)
             l[++tot]=Line(p[i],p[i+1]);
            for (int i=1;i<=tot;i++) l[i].slop=atan2(l[i].e.y-l[i].s.y,l[i].e.x-l[i].s.x); 
            double ll=0.0,rr=finf; 
            while (rr-ll>eps){
                double mid=(ll+rr)/2.0;
                if (check(mid)) ll=mid;
                else rr=mid;
            }
            printf("%.6f
    ",ll);
        }
    }
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  • 原文地址:https://www.cnblogs.com/qzqzgfy/p/5657481.html
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