Codeforces Round #403 B. The Meeting Place Cannot Be Changed（二分）

B. The Meeting Place Cannot Be Changed

Description

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.

Input

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.

Output

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 ∣a−b∣max(1,b)≤10−6 holds.

题意

在一维坐标轴上，有 n 个点 xi，每个点的最大移动速度为 vi 。问最短多少时间可以使得 n 个点在同一个位置相遇。

分析

读完题目就可以大致考虑通过二分枚举最短时间 t 来解决此题。

对于每个枚举的时间 t ，O(N) 处理每个点所能到达的最左最右区间，比较 n 个点的区间交是否存在即可。

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 60000 + 10;
const double eps = 1e-6;
int n, x
, v
;
bool jud(double t)
{
double l = 0, r = 10e9;
for(int i=1;i<=n;i++)
{
l = max(l, x[i]-v[i]*t);
r = min(r, x[i]+v[i]*t);
if(l>r) return false;
}
return true;
}
double solve()
{
double l = 0, r = 1e9, mid, ans;
while(r-l>eps)
{
mid = (l+r)/2.0;
if(jud(mid))
r = mid - eps,    ans = r;
else
l = mid + eps;
}
return ans;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&x[i]);
for(int i=1;i<=n;i++)
scanf("%d",&v[i]);
printf("%.10lf\n",solve());
}