http://acm.hdu.edu.cn/showproblem.php?pid=4717

大致题意：给出每一个点的坐标以及每一个点移动的速度和方向。

问在那一时刻点集中最远的距离在全部时刻的最远距离中最小。

比赛时一直以为是计算几何，和线段相交什么的有关。赛后队友说这是道三分，细致想了想确实是三分。试着画绘图发现它是一个凸性函数，存在一个最短距离。

然后三分时间就能够了。

#include <stdio.h>

#include <iostream>

#include <map>

#include <stack>

#include <vector>

#include <math.h>

#include <string.h>

#include <queue>

#include <string>

#include <stdlib.h>

#include <algorithm>

#define LL long long

#define _LL __int64

#define eps 1e-8

#define PI acos(-1.0)

using namespace std;

const int maxn = 310;

const int INF = 0x3f3f3f3f;

struct node

{

	double x,y,vx,vy;

}p[maxn],tp[maxn];

int n;

double mindis,time;

double dis(node a, node b)

{

	return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));

}

double cal(double t)

{

	for(int i = 1; i <= n; i++)

	{

		tp[i].x = p[i].x + t * p[i].vx;

		tp[i].y = p[i].y + t * p[i].vy;

	}

	double Max = -1.0;

	for(int i = 1; i < n; i++)

	{

		for(int j = i+1; j <= n; j++)

		{

			double ans = dis(tp[i], tp[j]);

			if(Max < ans)

				Max = ans;

		}

	}

	return Max;

}

void solve()

{

	double low = 0, high = INF, mid,midmid;

	while(high - low > eps)

	{

		mid = (high + low)/2;

		midmid = (mid + high)/2;

		double ans1 = cal(mid);

		double ans2 = cal(midmid);

		if(ans1 > ans2)

			low = mid;

		else

			high = midmid;

	}

	time = low;

	mindis = cal(low);

}

int main()

{

	int test;

	scanf("%d",&test);

	for(int item = 1; item <= test; item++)

	{

		scanf("%d",&n);

		for(int i = 1; i <= n; i++)

			scanf("%lf %lf %lf %lf",&p[i].x, &p[i].y, &p[i].vx,&p[i].vy);

		solve();

		printf("Case #%d: %.2lf %.2lf\n",item,time,mindis);

	}

	return 0;

}