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  • 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;
    }
    


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