cf1059D. Nature Reserve(三分)

題意

題目鏈接

Sol

欲哭無淚啊qwq。。。。昨晚一定是智息了qwq

說一個和標算不一樣做法吧。。

顯然\(x\)軸是可以三分的，半徑是可以二分的。

恭喜你獲得了一個TLE的做法。。

然後第二維的二分是沒有必要的，直接拿圓的標準方程推一下取個最大值就行了。。。。。昨晚沒想到qwq給數學老師丟臉了。。

#include<cstdio>
#include<cmath>
#include<algorithm> 
#define double long double 
using namespace std;
const double eps = 1e-7, INF = 1e18;
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0',c = getchar();
    return x * f;
}
int N, up, down;
double max(double a, double b) {return a > b ? a : b;}
double min(double a, double b) {return a < b ? a : b;}
struct Node {
    double x, y;
}a[MAXN];
int check(int x, int y) {
    if(x < 0 && y > 0) return 1;
    else return 0;
}
double mxr;
double sqr(double x) {
    return x * x;
}
double f(double x) {
    double mx = 0;
    for(int i = 1; i <= N; i++) 
        mx = max(mx, fabs((sqr(a[i].x - x) + sqr(a[i].y)) / (2.0 * a[i].y)));
    return mx;
}
      
int main() {
    N = read();
    double L = INF, R = -INF;
    for(int i = 1; i <= N; i++) {
        a[i].x = read(), a[i].y = read();
        up = min(up, a[i].y);
        mxr = max(a[i].y, mxr);
        L = min(a[i].x, L); 
        R = max(a[i].x, R);
    }
    if(check(up, mxr)) {puts("-1"); return 0;}
    mxr = INF;
    if(up < 0) for(int i = 1; i <= N; i++) a[i].y = -a[i].y;
    int tim = 100;
    while(tim--) {
        double x = (2 * L + R) / 3, y = (L + 2 * R) / 3;
        f(x) < f(y) ? R = y : L = x;
      //  printf("%Lf %Lf\n", f(x), f(y));
    }
    printf("%.10Lf", f(L));
    return 0;   
}