【二分，三分】Codeforces Round #403 The Meeting Place Cannot Be Changed

題意： 在x座標軸上，給出n個人的橫座標的位置和每個人行走的速度，問n個人在某個點集合最短要用的時間。

題解：

方法一：由於時間越大，大到一定程度一定能全部人集合，那麼二分時間。至於如何判斷時間是否符合，那個時間得到每個人能走的範圍，求範圍是否有交集即可。

方法二：由於時間隨 x 的變化函式是凹型函式，那麼可以三分 x 位置。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 60010;
const double esp = 1e-6;

struct asd
{
    double x,v;
}a[N];
int n;
bool slove(double t)
{
    double l = a[0].x - t*a[0].v;
    double r = a[0].x + t*a[0].v;
    for(int i = 1; i < n; i++)
    {
        double nextl = a[i].x - t*a[i].v;
        double nextr = a[i].x + t*a[i].v;
        if(nextl >= l && nextl <= r && nextr >= l && nextr <= r)
            l = nextl, r = nextr;
        else if(nextl >= l && nextl <= r)
            l = nextl;
        else if(nextr >= l && nextr <= r)
            r = nextr;
        else if(nextl <= l && nextr >= r)
            continue;
        else
            return false;
    }
    return true;
}
double Bsearch(double l,double r)
{
    double mid;
    while(fabs(l-r) > esp)
    {
        mid = l+(r-l)/2;
        if(slove(mid))
            r = mid;
        else
            l = mid;
    }
    return mid;
}
int main()
{
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> a[i].x;
    for(int i = 0; i < n; i++)
        cin >> a[i].v;
    //slove(2.0);
    double t = Bsearch(0,1000000000);
    printf("%.12lf\n",t);
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
const double INF = 0x3f3f3f3f;
const int N = 60010;
const double esp = 1e-6;

struct asd
{
    double x,v;
}a[N];
int n;
double slove(double f)
{
    double mx = 0;
    for(int i = 0; i < n; i++)
    {
        double t = fabs(a[i].x-f)/a[i].v;
        mx = max(mx,t);
    }
    return mx;
}
double sanBsearch(double l,double r)
{
    double midl,midr;
    while(fabs(l-r) > esp)
    {
        midl = l+(r-l)/3;
        midr = r+(l-r)/3;
        if(slove(midr) > slove(midl))
            r = midr;
        else
            l = midl;
    }
    return slove(midr);
}
int main()
{
    cin >> n;
    double mn = INF, mx = 0;
    for(int i = 0; i < n; i++)
    {
        cin >> a[i].x;
        mx = max(mx,a[i].x);
        mn = min(mn,a[i].x);
    }
    for(int i = 0; i < n; i++)
        cin >> a[i].v;
    //slove(2.0);
    double t = sanBsearch(mn,mx);
    printf("%.12lf\n",t);
    return 0;
}