POJ - 3525 E - Most Distant Point from the Sea(半平面交+二分)

POJ - 3525 E - Most Distant Point from the Sea
题意:给你一个日本, 日本中有一个地方, 离海的最近距离比其他地方的最近距离都大.要你求该地到海的最近距离.
也就是给你一个凸包, 在里面放一个最大的圆, 求出圆的半径
解法:
可以二分半径
每次将所有线段向圆心平移mid个单位长度
如果还存在核,则l=mid,否则r=mid
代码:

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1e3+5;
const double EPS=1e-9;
const double PI=acos(-1);
inline int sgn(double a){ return a < -EPS ? -1 : a > EPS; }
inline int cmp(double a, double b){ return sgn(a-b); }
int n,m,T;

struct Point;
struct Line;
typedef Point Vector;

struct Point{
    double x,y;
    Point(){}
    Point(double a, double b):x(a),y(b){}
    double len(){return sqrt(x*x+y*y);}
    Vector Normal(){//单位法向量逆时针旋转90
        return Vector(-y/len(),x/len());
    }
    void read(){scanf("%lf%lf",&x,&y);}
    Point operator+(Vector v){return {x+v.x,y+v.y};}
    Vector operator-(Point p){return {x-p.x,y-p.y};}
    double operator^(Vector v){return x*v.y-y*v.x;}//叉乘
    double operator*(Vector v){return x*v.x+y*v.y;}//点乘
    Vector operator*(double d){return {x*d,y*d};}
    Vector operator/(double d){return {x/d,y/d};}
    bool operator==(Point p){return cmp(x,p.x)==0&&cmp(y,p.y)==0;}
    bool operator<(Point p){if(cmp(x,p.x)==0) return y<p.y;return x<p.x;}
};
struct Line{
    Line(){}
    Line(Point a,Point b){s=a;e=b;ang=atan2(e.y-s.y,e.x-s.x);}
    Point point(double t){return s + (e - s)*t;}//返回点P = v + (p - v)*t
    Point operator &(Line &b){
        double t=((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        return s+(e-s)*t;
    }
    Point s,e;
    double ang;
};


Line Q[1510];
Point p[1510];
Line L[1510];
bool HPIcmp(Line a,Line b) {
    if (fabs(a.ang - b.ang) > EPS)return a.ang < b.ang;
    else return ((a.s - b.s) ^ (b.e - b.s)) < 0;
}
bool 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++){ // 去掉右边的线段，保留左边的线段。
        if (fabs(line[i].ang - line[i - 1].ang) > 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 false;
        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].s)) > EPS)
        tail--;
    return tail > head + 1;
}
double CalcArea(Point p[],int n) {
    double ans = 0;
    for (int i = 0; i < n; i++) {
        ans += (p[i] ^ p[(i + 1) % n]) / 2;
    }
    return ans;
}
Line tmp[MAXN];
void solve(){
    double l=0,r=10000;
    while(cmp(l,r)){
        double mid=(l+r)/2;
        for(int i=0;i<n;i++){
            Vector v=L[i].e-L[i].s;
            v=v.Normal();//垂直线段指向圆心的单位法向量
            tmp[i]={L[i].s+v*mid,L[i].e+v*mid};
        }
        int resn=0;
        if(HPI(tmp,n,p,resn)) l=mid;
        else r=mid;
    }
    printf("%.6f\n",l);
}
void init(){
    for(int i=0;i<n;i++)
        p[i].read();
    if(CalcArea(p,n)<0)
        reverse(p,p+n);
    for(int i=0;i<n;i++)
        L[i]={p[i],p[(i+1)%n]};
}
int main(){
    while(~scanf("%d",&n)&&n){
        init();
        solve();
    }
    return 0;
}

来源：CSDN

作者：0ng

链接：https://blog.csdn.net/weixin_43804251/article/details/104136861