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

题意： 在x坐标轴上，给出n个人的横坐标的位置和每个人行走的速度，问n个人在某个点集合最短要用的时间。

题解：

方法一：由于时间越大，大到一定程度一定能全部人集合，那么二分时间。至于如何判断时间是否符合，那个时间得到每个人能走的范围，求范围是否有交集即可。

方法二：由于时间随 x 的变化函数是凹型函数，那么可以三分 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
;
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
;
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;
}