hdu 4717(三分求极值)
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; }