hdu 4717 The Moving Points(三分+计算几何)

时间:2023-03-09 03:39:15
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,

hdu 4717 The Moving Points(三分+计算几何)

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-;
const double PI = acos(-1.0);
const double INF = 1000000000000000.000; struct Point{
double x,y;
Point(double x=, double y=) : 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 ;
else return x < ? - : ;
}
bool operator == (const Point& a, const Point& b){
return dcmp(a.x - b.x) == && dcmp(a.y - b.y) == ;
} ///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 = ; Point P[maxn];
Vector V[maxn];
int N; double calMax(double t){
double ret = ;
for(int i=;i<=N;i++)
for(int j=i+;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=;cas<=T;cas++){
cin>>N;
for(int i=;i<=N;i++){
P[i] = read_point();
V[i] = read_point();
} double Lt=;
double Rt=1e7; double M1t,M1w;
double M2t,M2w;
while(dcmp(Rt-Lt)>){
M1t = Lt+(Rt-Lt)/;
M1w = calMax(M1t); M2t = Lt+(Rt-Lt)/*;
M2w = calMax(M2t); if(dcmp(M1w-M2w)>=){
Lt = M1t+eps;
}
else{
Rt = M2t-eps;
}
}
printf("Case #%d: %.2lf %.2lf\n",cas,Lt,calMax(Lt));
}
}