• hdu 4717 The Moving Points(三分+计算几何)


    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4717

    说明下为啥满足三分:

    设y=f(x) (x>0)表示任意两个点的距离随时间x的增长,距离y的变化。则f(x)函数单调性有两种:1.先单减,后单增。2.一直单增。 

    设y=m(x) (x>0)表示随时间x的增长,所有点的最大距离y的变化。即m(x)是所有点对构成的f(x)图像取最上面的部分。则m(x)的单调性也只有两种可能:1.先单减,后单增。2.一直单增。 这个地方的证明可以这样:假如时刻t1到时刻t2最大值取得是函数f1(x)的图像,在时刻t2到时刻t3取得是f2(x)的图像,

    那么由图可以看出f2(x)的斜率大于f1(x)的斜率

    可以归纳出m(x)函数的斜率是递增。那么单调性就可以知道了。

    m(x)有了上面的性质,就可以有三分了。

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<iostream>
    #include<algorithm>
    #include<queue>
    using namespace std;
    
    const double eps = 1e-8;
    const double PI = acos(-1.0);
    const double INF = 1000000000000000.000;
    
    struct Point{
        double x,y;
        Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
    };
    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 * (double p,Vector A){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;
        else              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;
    }
    
    ///向量(x,y)的极角用atan2(y,x);
    inline double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
    inline double Length(Vector A)    { return sqrt(Dot(A,A)); }
    inline double Angle(Vector A, Vector B)  { return acos(Dot(A,B) / Length(A) / Length(B)); }
    
    Point read_point(){
        Point A;
        scanf("%lf %lf",&A.x,&A.y);
        return A;
    }
    
    /*************************************分 割 线*****************************************/
    const int maxn = 305;
    
    Point P[maxn];
    Vector V[maxn];
    int N;
    
    double calMax(double t){
        double ret = 0;
        for(int i=1;i<=N;i++)
            for(int j=i+1;j<=N;j++){
                double len = Length(P[i]+t*V[i]-(P[j]+t*V[j]));
                ret = max(ret,len);
        }
        return  ret;
    }
    
    int main()
    {
        //freopen("E:\acm\input.txt","r",stdin);
        int T;
        cin>>T;
        for(int cas=1;cas<=T;cas++){
            cin>>N;
            for(int i=1;i<=N;i++){
                P[i] = read_point();
                V[i] = read_point();
            }
    
            double Lt=0;
            double Rt=1e7;
    
            double M1t,M1w;
            double M2t,M2w;
            while(dcmp(Rt-Lt)>0){
                M1t = Lt+(Rt-Lt)/3;
                M1w = calMax(M1t);
    
                M2t = Lt+(Rt-Lt)/3*2;
                M2w = calMax(M2t);
    
                if(dcmp(M1w-M2w)>=0){
                    Lt = M1t+eps;
                }
                else{
                    Rt = M2t-eps;
                }
            }
            printf("Case #%d: %.2lf %.2lf
    ",cas,Lt,calMax(Lt));
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/acmdeweilai/p/3315769.html
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