The Moving Points HDU - 4717

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.

InputThe 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 ^thpoint, 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).OutputFor 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



这题也是三分的入门题（水题）
给你N个坐标，和每秒在特定方向的移动速度。
求出最大距离的最小值。（是不是和我上一篇写的，题目非常的相似）
其实两点之间的距离与时间t的关系刚好是一个二次函数。
其实又是构造一个由最大值组成的二次函数求出这个新的二次函数的最小值

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
struct node
{
    double x,y,vx,vy;
}a[510];
double f(node a,node b ,double t)
{
    return sqrt((a.x+a.vx*t-b.x-b.vx*t)*(a.x+a.vx*t-b.x-b.vx*t)+(a.y+a.vy*t-b.y-b.vy*t)*(a.y+a.vy*t-b.y-b.vy*t));
}
double check(double x)
{
    double maxn=0;
    for (int i=0 ;i<n ;i++){
        for (int j=i+1 ;j<n ;j++){
            maxn=max(maxn,f(a[i],a[j],x));
        }
    }
    return maxn;
}
int main() {
    int t,cas=1;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for (int i=0 ;i<n ;i++){
            scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&a[i].vx,&a[i].vy);
        }
        double l=0,r=1000000;
        while(r-l>1e-9){
            double lmid=l+(r-l)/3;
            double rmid=r-(r-l)/3;
            if (check(lmid)>check(rmid)) l=lmid;
            else r=rmid;
        }
        printf("Case #%d: %.2lf %.2lf\n",cas++,l,check(l));
    }
    return 0;
}