The Meeting Place Cannot Be Changed
题面:
The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction.
At some points on the road there are n friends, and i-th of them is standing at the point xi meters and can move with any speed no greater than vi meters per second in any of the two directions along the road: south or north.
You are to compute the minimum time needed to gather all the n friends at some point on the road. Note that the point they meet at doesn't need to have integer coordinate.
The first line contains single integer n (2 ≤ n ≤ 60 000) — the number of friends.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) — the current coordinates of the friends, in meters.
The third line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109) — the maximum speeds of the friends, in meters per second.
Print the minimum time (in seconds) needed for all the n friends to meet at some point on the road.
Your answer will be considered correct, if its absolute or relative error isn't greater than 10 - 6. Formally, let your answer be a, while jury's answer be b. Your answer will be considered correct if holds.
很容易就可以联想到可以用二分来解决,这里选择二分时间来枚举区间,check函数判断对于每一个时间是否可以保证所有人能到达得一个区间,只要存在一个人无法到达左后得小区间就返回false
但是写代码得过程中遇到一些精度问题,自己得二分和chenk函数应该都是没问题,但是怎么都过不了样例。后来东巨发现我写的double类型判断大小的函数函数应该加上精度考虑,就对了。xwdtql!
贴代码
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-7 #define INF 1e18 #define maxn 60005 int n; double xi[maxn],sp[maxn]; double Max(double a, double b) { return (a - b > eps) ? a : b ; //这里如果写成 a - b > 0 ? a : b 就不行 } double Min(double a, double b) { return (a - b > eps) ? b : a; } bool check(double t) { double lp = xi[1] - sp[1] * t, rp = xi[1] + sp[1] * t; for (int i = 1; i <= n; ++i) { double a, b; a = xi[i] - sp[i] * t; b = xi[i] + sp[i] * t; lp = Max(lp, a); rp = Min(rp, b); } if (lp <= rp) return true; return false; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%lf",&xi[i]); for(int i=1;i<=n;i++) scanf("%lf",&sp[i]); double l=0,r=1e9,ans; while(l<=r) { double mid=(l+r)/2.0; if(check(mid)) r=mid-eps,ans=mid; else l=mid+eps; } printf("%.7lf",ans); return 0; }