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)); } }