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

思路：三分时间求极小值。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const int MAX_N = (300 + 30);

const double eps = 1e-5;

struct Point {

    double x, y;

    double vx, vy;

} point[MAX_N];

int N;

double ans;

double getDist(int i, int j, double t_time)

{

    double x1 = (point[i].x + t_time * point[i].vx);

    double y1 = (point[i].y + t_time * point[i].vy);

    double x2 = (point[j].x + t_time * point[j].vx);

    double y2 = (point[j].y + t_time * point[j].vy);

    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));

}

double check(double t_time)

{

    double res = 0.0;

    for (int i = 1; i <= N; ++i) {

        for (int j = i + 1; j <= N; ++j) {

            res = max(res, getDist(i, j, t_time));

        }

    }

    ans = min(ans, res);

    return res;

}

int main()

{

    int Cas, t = 1;

    scanf("%d", &Cas);

    while (Cas--) {

        scanf("%d", &N);

        for (int i = 1; i <= N; ++i) {

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

        }

        if (N == 1) {

            printf("%.2f %.2f\n", 0, 0);

            continue;

        }

        ans  = 1e18;

        double low = 0.0, high = 1e8, mid, mmid;

        while (low + eps < high) {

            mid = (low + high) / 2.0;

            mmid = (mid + high) / 2.0;

            if (check(mid) <= check(mmid)) {

                high = mmid;

            } else

                low = mid;

        }

        printf("Case #%d: %.2f %.2f\n", t++, low, ans);

    }

    return 0;

}