B. The Meeting Place Cannot Be Changed
time limit per test5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 holds.
Examples
input
3
7 1 3
1 2 1
output
2.000000000000
input
4
5 10 3 2
2 3 2 4
output
1.400000000000
Note
In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.
題意：在x軸上有很多人，這些人都有對應的最大移動速度，想要使所有人運動至同一點，求出最小運動時間。
思路：二分枚舉時間，對於枚舉到的中間值mid，假設所有人往右走，求出所有人往右走最左邊的位置，接着判斷所有人是否能在mid時間內到達該點（事實上只需要判斷位於該點右側的人是否能在mid時間內到達該點）
代碼
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
const int maxn=60005;
const int INF=1000000005;
struct node
{
int x;//座標
int v;//速度
} num[maxn];
int N;
bool is_ok(double T)
{
//找到當前時間T下，所有人向有運動能到達的最左邊位置
double point=INF*1.0;
for(int i=1; i<=N; i++)
point=min(point,num[i].x*1.0+num[i].v*T*1.0);
for(int i=1; i<=N; i++)
{
if(num[i].x*1.0>point)//需要向左跑的人
{
if(num[i].x*1.0-num[i].v*T*1.0>point)
{
return false;
}
}
}
return true;
}
int main()
{
scanf("%d",&N);
for(int i=1; i<=N; i++)
scanf("%d",&num[i].x);
for(int i=1; i<=N; i++)
scanf("%d",&num[i].v);
double left=0,right=INF;
while((right-left)>0.0000001)
{
double mid=(left+right)/2.0;
if(is_ok(mid)==false)
left=mid;
else
right=mid;
}
printf("%lf\n",left);
return 0;
}