每两个点的距离函数是先递减再递增或者是递增,想到三分。
所有的两点距离的图像都是二次函数开口向上,所以两两合并到最后只剩一个图像的时候还是二次函数开口向上,三分就可以了。
#include <bits/stdc++.h> using namespace std; const int maxn = 350; const double EPS = 1e-8; const double INF = 0x3f3f3f3f; struct Point { double x,y; double vx,vy; } p[maxn]; int n; double dis(Point p1,Point p2,double t) { double x1 = p1.x + p1.vx*t; double y1 = p1.y + p1.vy*t; double x2 = p2.x + p2.vx*t; double y2 = p2.y + p2.vy*t; return sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) ); } double cal(double t) { double Max = 0; for(int i = 0 ; i < n ; i++) { for(int j = i + 1 ; j < n ; j++) { Max = max(Max, dis(p[i],p[j],t)); } } return Max; } double ThreeCut(double left,double right) { double mid,midmid; while(fabs(left - right) > EPS) { mid = (left + right) / 2; midmid = (mid + right) / 2; if(cal(mid) > cal(midmid)) { left = mid; } else { right = midmid; } } return midmid; } int main() { ios::sync_with_stdio(false); int T; cin >> T; int kase = 1; while(T--) { cin >> n; for(int i = 0 ; i < n ; i++) { cin >> p[i].x >> p[i].y >> p[i].vx >> p[i].vy; } printf("Case #%d: %.2lf %.2lf\n",kase++,ThreeCut(0,INF),cal(ThreeCut(0,INF))); } return 0; }