题意:给出$n$个坐标$x_i$,$n$个速度$v_i$问使他们相遇的最短时间是多少。
思路:首先可肯定最终相遇位置必定在区间$[0,max(x_i)]$中,二分最终位置,判断左右部分各自所花时间,缩小范围即可。
/** @Date : 2017-05-09 22:07:43 * @FileName: 782B.cpp * @Platform: Windows * @Author : Lweleth (SoundEarlf@gmail.com) * @Link : https://github.com/Lweleth * @Version : $Id$ */ #include <bits/stdc++.h> #define LL long long #define PII pair #define MP(x, y) make_pair((x),(y)) #define fi first #define se second #define PB(x) push_back((x)) #define MMG(x) memset((x), -1,sizeof(x)) #define MMF(x) memset((x),0,sizeof(x)) #define MMI(x) memset((x), INF, sizeof(x)) using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5+20; const double eps = 1e-6; double p[60010]; double v[60010]; double tim; int n; int check(double x) { double lt = 0, rt = 0; for(int i = 0; i < n; i++) { if(p[i] < x) lt = max(lt, (x - p[i]) / v[i]); else if(p[i] > x) rt = max(rt, (p[i] - x) / v[i]); } tim = max(lt, rt); return (lt >= rt); } int main() { while(cin >> n) { double ma = 0; for(int i = 0; i < n; i++) scanf("%lf", p + i), ma = max(ma, p[i]); for(int i = 0; i < n; i++) scanf("%lf", v + i); double l = 0, r = ma; while(r - l >= eps) { double mid = (r + l) / 2.0000; if(check(mid)) r = mid; else l = mid; //cout << mid << endl; } printf("%.10lf ", tim); } return 0; }