[HDU 4855 Goddess] 同号单调函数求和+三分

题目

http://acm.hdu.edu.cn/showproblem.php?pid=4855

分析

三分

首先可以容易发现对于每个圆，当射线从圆心极角向切线极角移动时，长度单调递减

然后把所有圆的圆心极角和切线极角排序，分成很多区间，每个区间内是同号单调函数的和，得到的函数为单峰函数，可以三分

因为问题保证圆不会覆盖原点，所以判断圆是否与射线相交可以通过判断圆心在射线的哪一侧来解决

代码

/**************************************************
 *        Problem:  HDU 4855 (Xi'an Invitation 2014 Problem I)
 *         Author:  clavichord93
 *          State:  Accepted
 **************************************************/

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define sqr(a) ((a) * (a))
using namespace std;
const double DINF = 1e30;
const double PI = acos(-1.0);
const double EPS = 1e-13;

inline int sgn(double a) {
    return a > EPS ? 1 : (a < -EPS ? -1 : 0);
}

const int MAX_N = 205;

struct Circle {
    double x, y, r;
    void read() {
        scanf("%lf %lf %lf", &x, &y, &r);
    }
};

int n;
Circle c[MAX_N];

int cntAng;
double ang[MAX_N * 3];

double calc(double ang) {
    double A = sin(ang);
    double B = -cos(ang);
    double _A = cos(ang);
    double _B = sin(ang);
    double ret = 0;
    for (int i = 0; i < n; i++) {
        double dist = fabs(A * c[i].x + B * c[i].y);
        double tag = _A * c[i].x + _B * c[i].y;
        if (sgn(dist - c[i].r) < 0 && sgn(tag) > 0) {
            double length = 2.0 * sqrt(sqr(c[i].r) - sqr(dist));
            ret += length;
        }
    }
    return ret;
}

int main() {
    #ifdef LOCAL_JUDGE
    freopen("in.txt", "r", stdin);
    #endif
    while (scanf("%d", &n) != EOF) {
        cntAng = 0;
        ang[cntAng++] = -PI;
        ang[cntAng++] = PI;
        for (int i = 0; i < n; i++) {
            c[i].read();
            double angCenter = atan2(c[i].y, c[i].x);
            double dist = sqrt(sqr(c[i].x) + sqr(c[i].y));
            double alpha = asin(c[i].r / dist);
            double angLeft = angCenter - alpha;
            double angRight = angCenter + alpha;
            if (sgn(angLeft + PI) < 0) {
                angLeft += 2.0 * PI;
            }
            if (sgn(angRight - PI) > 0) {
                angRight -= 2.0 * PI;
            }
            ang[cntAng++] = angLeft;
            ang[cntAng++] = angCenter;
            ang[cntAng++] = angRight;
        }
        sort(ang, ang + cntAng);
        cntAng = unique(ang, ang + cntAng) - ang;
        double ans = 0;
        for (int i = 0; i < cntAng - 1; i++) {
            double l = ang[i], r = ang[i + 1];
            double res = 0;
            while (l <= r) {
                double length = (r - l) / 3;
                double x1 = l + length;
                double x2 = l + length * 2;
                double f1 = calc(x1);
                double f2 = calc(x2);
                if (sgn(f1 - f2) >= 0) {
                    res = f1;
                    r = x2 - EPS;
                }
                else {
                    res = f2;
                    l = x1 + EPS;
                }
            }
            if (sgn(ans - res) < 0) {
                ans = res;
            }
        }
        printf("%.10f\n", ans);
    }

    return 0;
}