UVALive 3890 Most Distant Point from the Sea(凸包最大内接园)

一个n个点的凸多边形,求多边形中离多边形边界最远的距离。实际上就是求凸包最大内接圆的半径。

利用半平面交求解,每次二分枚举半径d,然后将凸包每条边所代表的半平面沿其垂直单位法向量平移d,看所有平移后的半平面的交集是否为空。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<bitset>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define eps 1e-10
using namespace std;

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

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point& b)
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

int dcmp(double x)
{
    if(fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b)
{
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
double Area2(Vector A, Vector B, Vector C) { return Cross(B-A, C-A); }
Vector Normal(Vector a) //a向量的垂直法向量
{
     return Vector(-a.y/Length(a), a.x/Length(a));
}

struct Line
{
    Point p;
    Vector v;
    double ang;
    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在半平面的左边
bool onLeft(Line L, Point p) { return Cross(L.v, p-L.p) > 0; }
//直线交点
Point GetIntersection(Line a, 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 int maxn = 200;
Point p[maxn], poly[maxn];
Line L[maxn];
Vector v[maxn], v2[maxn];
int n;

//半平面交
Point pp[maxn];
Line qq[maxn];
int HalfplaneIntersection(Line* L, int n, Point* poly)
{
    sort(L, L+n);
    int first, last;

    qq[first=last=0] = L[0];
    FF(i, 1, n)
    {
        while(first < last && !onLeft(L[i], pp[last-1])) last--;
        while(first < last && !onLeft(L[i], pp[first])) first++;
        qq[++last] = L[i];
        if(fabs(Cross(qq[last].v, qq[last-1].v)) < eps)
        {
            last--;
            if(onLeft(qq[last], L[i].p)) qq[last] = L[i];
        }
        if(first < last) pp[last-1] = GetIntersection(qq[last-1], qq[last]);
    }
    while(first < last && !onLeft(qq[first], pp[last-1])) last--;
    if(last-first <= 1) return 0;
    pp[last] = GetIntersection(qq[last], qq[first]);

    int m = 0;
    FF(i, first, last+1) poly[m++] = pp[i];
    return m;
}


int main()
{
    while(scanf("%d", &n), n)
    {
        REP(i, n) scanf("%lf%lf", &p[i].x, &p[i].y);
        REP(i, n)
        {
            v[i] = p[(i+1)%n]-p[i];
            v2[i] = Normal(v[i]);
        }
        double l=0, r=20000, mid;
        while(r - l > eps)
        {
            mid = (l+r) / 2.0;
            REP(i, n) L[i] = Line(p[i]+v2[i]*mid, v[i]);
            int m = HalfplaneIntersection(L, n, poly);
            if(!m) r=mid; else l=mid;
        }
        printf("%.6f
", l);
    }
    return 0;
}


原文地址:https://www.cnblogs.com/riskyer/p/3293908.html