CodeForces - 782B The Meeting Place Cannot Be Changed(精度二分)

题意：在一维坐标轴上，给定n个点的坐标以及他们的最大移动速度，问他们能聚到某一点处的最短时间。

分析：

1、二分枚举最短时间即可。

2、通过检查当前时间下，各点的最大移动范围之间是否有交集，不断缩小搜索范围。

3、相当于二分枚举找右临界线，符合要求的点都在右边。

4、通过给二分一个查找次数的上界，eg：k=200，而不是dcmp(l, r)<=0，后者会tle。

5、检查是否有交集，就是以第一个点的最大移动区间为标准，不断与其他区间取交集并更新再继续比较。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 60000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN], v[MAXN];
int n;
bool judge(double t){
    double tmpl = (double)a[0] - (double)v[0] * t;
    double tmpr = (double)a[0] + (double)v[0] * t;
    for(int i = 1; i < n; ++i){
        double l = (double)a[i] - (double)v[i] * t;
        double r = (double)a[i] + (double)v[i] * t;
        if(dcmp(tmpl, r) > 0 || dcmp(tmpr, l) < 0) return false;
        if(dcmp(l, tmpl) > 0) tmpl = l;
        if(dcmp(r, tmpr) < 0) tmpr = r;
    }
    return true;
}
double solve(){
    double l = 0, r = (double)1e9;
    int k = 200;
    while(k--){
        double mid = (l + r) / 2;
        if(judge(mid)) r = mid;
        else l = mid + eps;
    }
    return r;
}
int main(){;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        scanf("%d", &a[i]);
    }
    for(int i = 0; i < n; ++i){
        scanf("%d", &v[i]);
    }
    printf("%.12lf\n", solve());
    return 0;
}