uvalive 3890 Most Distant Point from the Sea

题意：求一个凸多边形中一点到边的最大距离。

思路：转换成在多边形内部，到每边距离为d的直线所围成的内多边形是否存在。也就是，二分距离+半平面交。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
#define up(i,x,y) for(i=x;i<=y;i++)
#define w(a) while(a)
const double inf=0x3f3f3f3f;
const int N = 4010;
const double eps = 5*1e-13;
const double PI = acos(-1.0);
using namespace std;

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator + (const Vector& A, const Vector& B)
{
    return Vector(A.x+B.x, A.y+B.y);
}
Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}
Vector operator * (const Vector& A, double p)
{
    return Vector(A.x*p, A.y*p);
}
double Dot(const Vector& A, const Vector& B)
{
    return A.x*B.x + A.y*B.y;
}
double Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}
double Length(const Vector& A)
{
    return sqrt(Dot(A, A));
}
Vector Normal(const Vector& A)
{
    double L = Length(A);
    return Vector(-A.y/L, A.x/L);
}

double PolygonArea(vector<Point> p)
{
    int n = p.size();
    double area = 0;
    for(int i = 1; i < n-1; i++)
        area += Cross(p[i]-p[0], p[i+1]-p[0]);
    return area/2;
}

// 有向直线。它的左边就是对应的半平面
struct Line
{
    Point P;    // 直线上任意一点
    Vector v;   // 方向向量
    double ang; // 极角，即从x正半轴旋转到向量v所需要的角（弧度）
    Line() {}
    Line(Point P, Vector v):P(P),v(v)
    {
        ang = atan2(v.y, v.x);
    }
    bool operator < (const Line& L) const
    {
        return ang < L.ang;
    }
};

// 点p在有向直线L的左边（线上不算）
bool OnLeft(const Line& L, const Point& p)
{
    return Cross(L.v, p-L.P) > 0;
}

// 二直线交点，假定交点惟一存在
Point GetLineIntersection(const Line& a, const Line& b)
{
    Vector u = a.P-b.P;
    double t = Cross(b.v, u) / Cross(a.v, b.v);
    return a.P+a.v*t;
}

const double INF = 1e8;

// 半平面交主过程
vector<Point> HalfplaneIntersection(vector<Line> L)
{
    int n = L.size();
    sort(L.begin(), L.end()); // 按极角排序

    int first, last;         // 双端队列的第一个元素和最后一个元素的下标
    vector<Point> p(n);      // p[i]为q[i]和q[i+1]的交点
    vector<Line> q(n);       // 双端队列
    vector<Point> ans;       // 结果

    q[first=last=0] = L[0];  // 双端队列初始化为只有一个半平面L[0]
    for(int i = 1; i < n; i++)
    {
        while(first < last && !OnLeft(L[i], p[last-1])) last--;
        while(first < last && !OnLeft(L[i], p[first])) first++;
        q[++last] = L[i];
        if(fabs(Cross(q[last].v, q[last-1].v)) < eps)   // 两向量平行且同向，取内侧的一个
        {
            last--;
            if(OnLeft(q[last], L[i].P)) q[last] = L[i];
        }
        if(first < last) p[last-1] = GetLineIntersection(q[last-1], q[last]);
    }
    while(first < last && !OnLeft(q[first], p[last-1])) last--; // 删除无用平面
    if(last - first <= 1) return ans; // 空集
    p[last] = GetLineIntersection(q[last], q[first]); // 计算首尾两个半平面的交点

    // 从deque复制到输出中
    for(int i = first; i <= last; i++) ans.push_back(p[i]);
    return ans;
}

int main()
{
    int n;
    while(scanf("%d", &n) == 1 && n)
    {
        vector<Vector> p, v, normal;
        int m, x, y;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &x, &y);
            p.push_back(Point(x,y));
        }
        if(PolygonArea(p) < 0) reverse(p.begin(), p.end());

        for(int i = 0; i < n; i++)
        {
            v.push_back(p[(i+1)%n]-p[i]);
            normal.push_back(Normal(v[i]));
        }

        double left = 0, right = 20000;
        while(right-left > 1e-6)
        {
            vector<Line> L;
            double mid = left+(right-left)/2;
            for(int i = 0; i < n; i++) L.push_back(Line(p[i]+normal[i]*mid, v[i]));//平移直线
            vector<Point> poly = HalfplaneIntersection(L);
            if(poly.empty()) right = mid;
            else left = mid;
        }
        printf("%.6lf
", left);
    }
    return 0;
}