F. Two Points
time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputThere are two points (x1, y1) and (x2, y2) on the plane. They move with the velocities (vx1, vy1) and (vx2, vy2). Find the minimal distance between them ever in future.
Input
The first line contains four space-separated integers x1, y1, x2, y2 ( - 104 ≤ x1, y1, x2, y2 ≤ 104) — the coordinates of the points.
The second line contains four space-separated integers vx1, vy1, vx2, vy2 ( - 104 ≤ vx1, vy1, vx2, vy2 ≤ 104) — the velocities of the points.
Output
Output a real number d — the minimal distance between the points. Absolute or relative error of the answer should be less than 10 - 6.
Examples
Input
1 1 2 2
0 0 -1 0
Output
1.000000000000000
Input
1 1 2 2
0 0 1 0
Output
1.414213562373095
题意：给你两个点 以及这两个点的x，y两个方向的速度 问运动过程中两个点间的最小距离
题解：第一次三分处理 类比二分 二分是针对单调函数的数值确定
三分不仅适用于单调函数，而且适用于凸函数
运动的两点间的距离的图像是不确定的 三分处理..
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<queue> 6 #include<stack> 7 #include<cmath> 8 using namespace std; 9 double x1,x2,y1,y2; 10 double vx1,vx2,vy1,vy2; 11 double f(double t) 12 { 13 double xx1,xx2,yy1,yy2; 14 xx1=x1+t*vx1; 15 xx2=x2+t*vx2; 16 yy1=y1+t*vy1; 17 yy2=y2+t*vy2; 18 return sqrt((xx1-xx2)*(xx1-xx2)+(yy1-yy2)*(yy1-yy2)); 19 } 20 int main() 21 { 22 scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2); 23 scanf("%lf %lf %lf %lf",&vx1,&vy1,&vx2,&vy2); 24 double l=0,r=1e9,m1,m2; 25 for(int i=0;i<200;i++) 26 { 27 m1=l+(r-l)/3; 28 m2=r-(r-l)/3; 29 if(f(m1)<f(m2)) 30 r=m2; 31 else 32 l=m1; 33 } 34 printf("%f ",f(l)); 35 return 0; 36 }