标签：nbsp time include measure rom tin input down algorithm
题意：
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.
3
7 1 3
1 2 1
2.000000000000
4
5 10 3 2
2 3 2 4
1.400000000000
思路：
二分，注意控制精度。
实现：
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <iomanip> 5 using namespace std; 6 struct node 7 { 8 double pos; 9 double speed; 10 }; 11 node a[60005]; 12 int n; 13 double minn, maxn; 14 bool check(double t) 15 { 16 minn = a[0].pos - t * a[0].speed; 17 maxn = a[0].pos + t * a[0].speed; 18 for (int i = 1; i < n; i++) 19 { 20 double tmp_l = a[i].pos - t * a[i].speed; 21 double tmp_r = a[i].pos + t * a[i].speed; 22 minn = max(minn, tmp_l); 23 maxn = min(maxn, tmp_r); 24 } 25 return maxn - minn >= 1e-7; 26 } 27 28 double solve() 29 { 30 double l = 0.0, r = 1000000005, res = 1000000005; 31 for (int i = 0; i < 10000; i++) 32 { 33 double mid = (l + r) / 2.0; 34 if (check(mid)) 35 { 36 r = mid; 37 res = mid; 38 } 39 else 40 { 41 l = mid; 42 } 43 } 44 return res; 45 } 46 47 int main() 48 { 49 cin >> n; 50 for (int i = 0; i < n; i++) 51 { 52 cin >> a[i].pos; 53 } 54 for (int i = 0; i < n; i++) 55 { 56 cin >> a[i].speed; 57 } 58 cout << setprecision(7) << solve() << endl; 59 return 0; 60 }
CF782B The Meeting Place Cannot Be Changed
标签：nbsp time include measure rom tin input down algorithm
原文地址：http://www.cnblogs.com/wangyiming/p/6516537.html
