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来源：互联网发布：重复犯错歌词意思知乎编辑：程序博客网时间：2024/06/07 06:21

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：在一个平面上有很多运动的点，问那一个时刻，所有点的最大距离最小。

先减后增因为绝对不会先增后减就是下凸模型
三分法

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int n;const double les=1e-6;struct node{    double x,y,vx,vy;}a[500];double dis(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 maxx=0;    for(int i=0;i<n;i++)    {        for(int j=i+1;j<n;j++)        {            maxx=max(maxx,dis(a[i],a[j],x));        }    }    return maxx;}int main(){    int T,t=1;scanf("%d",&T);    while(T--){        double maxx=-1;        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=1e6;        while(l+les<=r){            double mid=(l+r)*0.5;            double midd=(mid+r)*0.5;            if(check(mid)<check(midd)) r=midd;            else l=mid;        }        printf("Case #%d: ",t++);        if(n==1) printf("0.00 0.00\n");        else printf("%.2f %.2f\n",l,check(l) );    }}

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