HDU 4717 The Moving Points (三分 模板)

There are N points in total. Every point moves in certain direction and certain speed. We want to know at what time that the largest distance between any two points would be minimum. And also, we require you to calculate that
minimum distance. We guarantee that no two points will move in exactly same speed and direction.

Input

The rst line has a number T (T <= 10) , indicating the number of test cases. 

For each test case, first line has a single number N (N <= 300), which is the number of points. 

For next N lines, each come with four integers X i, Y i, VX i and VY i (-10 6 <= X i, Y i <= 10 6, -10 2 <= VX i , VY i <= 10 2), (X i, Y i) is the position of the i th point, and (VX i , VY i) is its speed with direction. That is to say, after 1 second, this
point will move to (X i + VX i , Y i + VY i).

Output

For test case X, output "Case #X: " first, then output two numbers, rounded to 0.01, as the answer of time and distance.

Sample Input

2

2

0 0 1 0

2 0 -1 0

2

0 0 1 0

2 1 -1 0

Sample Output

Case #1: 1.00 0.00

Case #2: 1.00 1.00

题意：给定n个点，然后这些点朝着给定的方向和给定速度走，在某个时间后，点停下来，可以求得此时任意两点之间的最大距离。问何时停止可以使得任意两点之间的最大距离最小。

思路：三分时间，判断条件是求出时间的情况下，任意两点之间的最大距离就好了。

三分：时间。

构造函数：运动之间的两点间的距离为先减小后增大（包含部分特殊的运动，比如速度相同，或者当前两点的距离已经为最小，之后变大。都在三分法的考虑范围之内）

所以构造函数为a>0,凸函数

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdio>
#include <functional> //小根堆特有头文件
typedef long long ll;
using namespace std;
const double eps=1e-8;
const int N=330;
int n;
double x
,y
,vx
,vy
;
double cal(double t)
{
double ret=0;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
{
double x1=x[i]+vx[i]*t;
double y1=y[i]+vy[i]*t;
double x2=x[j]+vx[j]*t;
double y2=y[j]+vy[j]*t;
ret=max(ret,sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
}
return ret;

}
double solve()
{
double l=0,r=1e10;
while(r-l>=eps)
{
double mid1=(l+r)/2;
double mid2=(mid1+r)/2;
if(cal(mid1)>=cal(mid2))
l=mid1;
else
r=mid2;
}
return l;
}
int main()
{
int icase,T=1;
scanf("%d",&icase);
while(icase--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%lf%lf%lf%lf",&x[i],&y[i],&vx[i],&vy[i]);
double t=solve();
printf("Case #%d: %.2lf %.2lf\n",T++,t,cal(t));
}
return 0;
}