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POJ3525 Most Distant Point from the Sea(半平面交)

2024-07-03 02:34:09   226人阅读

今天打算做两道半平面交，一题卡太久了，心都碎了。。。

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#pragma warning(disable:4996)
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
 
#define maxn 2500
#define eps 1e-7
 
int n;
 
int dcmp(double x){
    return (x > eps) - (x < -eps);
}
 
struct Point
{
    double x, y;
    Point(){}
    Point(double _x, double _y) :x(_x), y(_y){}
    Point operator + (const Point &b) const{
        return Point(x + b.x, y + b.y);
    }
    Point operator - (const Point &b) const{
        return Point(x - b.x, y - b.y);
 
    }
    Point operator *(double d) const{
        return Point(x*d, y*d);
    }
    Point operator /(double d) const{
        return Point(x / d, y / d);
    }
    double det(const Point &b) const{
        return x*b.y - y*b.x;
    }
    double dot(const Point &b) const{
        return x*b.x + y*b.y;
    }
    Point rot90(){
        return Point(-y, x);
    }
    Point norm(){
        double len=sqrt(this->dot(*this));
        return Point(x, y) / len;
    }
    void read(){
        scanf("%lf%lf", &x, &y);
    }
};
 
#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) (dcmp(cross(p1,p2,p3)))
 
Point isSS(Point p1, Point p2, Point q1, Point q2){
    double a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
    return (p1*a2 + p2*a1) / (a1 + a2);
}
 
struct Border
{
    Point p1, p2;
    double alpha;
    void setAlpha(){
        alpha = atan2(p2.y - p1.y, p2.x - p1.x);
    }
};
 
bool operator < (const Border &a,const Border &b) {
    int c = dcmp(a.alpha - b.alpha);
    if (c != 0) {
        return c == 1;
    }
    else {
        return crossOp(b.p1, b.p2, a.p1) > 0;
    }
}
 
bool operator == (const Border &a, const Border &b){
    return dcmp(a.alpha - b.alpha) == 0;
}
 
 
Point isBorder(const Border &a, const Border &b){
    return isSS(a.p1, a.p2, b.p1, b.p2);
}
 
Border border[maxn];
Border que[maxn];
int qh, qt;
// check函数判断的是新加的半平面和由a,b两个半平面产生的交点的方向，若在半平面的左侧返回True
bool check(const Border &a, const Border &b, const Border &me){
    Point is = isBorder(a, b);
    return crossOp(me.p1, me.p2, is) > 0;
}
 
bool convexIntersection()
{
    qh = qt = 0;
    sort(border, border + n);
    n = unique(border, border + n) - border;
    for (int i = 0; i < n; i++){
        Border cur = border[i];
        while (qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], cur)) --qt;
        while (qh + 1 < qt&&!check(que[qh], que[qh + 1], cur)) ++qh;
        que[qt++] = cur;
    }
    while (qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], que[qh])) --qt;
    while (qh + 1 < qt&&!check(que[qh], que[qh + 1], que[qt - 1])) ++qh;
    return qt - qh > 2;
}
 
Point ps[maxn];
 
bool judge(double x)
{
    for (int i = 0; i < n; i++){
        border[i].p1 = ps[i];
        border[i].p2 = ps[(i + 1) % n];
    }
    for (int i = 0; i < n; i++){
        Point vec = border[i].p2 - border[i].p1;
        vec=vec.rot90().norm();
        vec = vec*x;
        border[i].p1 = border[i].p1 + vec;
        border[i].p2 = border[i].p2 + vec;
        border[i].setAlpha();
    }
    return convexIntersection();
}
 
int main()
{
    while (cin>>n&&n)
    {
        for (int i = 0; i < n; i++){
            ps[i].read();
        }
        double l=0, r=100000000;
        while (dcmp(r-l)>0){
            double mid = (l + r) / 2;
            if (judge(mid)) l = mid;
            else r = mid;
        }
        printf("%.6lf\n", l);
    }
    return 0;
}

 


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