HDU 3264 Open-air shopping malls（圆相交面积+二分）

ACM

题目地址：HDU 3264 Open-air shopping malls

题意： 
给出一些圆，选择其中一个圆的圆心为圆心，然后画一个大圆，要求大圆最少覆盖每个圆的一半面积。求最小面积。

分析： 
枚举每个点，用二分求出需要的圆，更新最小值即可。 
其中用到了圆相交面积，可以参考这题： POJ 2546 Circular Area（两个圆相交面积）

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        3264.cpp*  Create Date: 2014-08-12 20:25:57*  Descripton:  Geometry, the insection of two round, binary*/#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 21;const double PI = acos(-1.0);const double EPS = 1e-8;int t, n;struct Round {    double x, y;    double r;} r[N], ori;double dis(const Round &a, const Round &b) {    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}double insection_of_rounds(Round &a, Round &b) {    double d = dis(a, b);    if (d >= a.r + b.r)        return 0;    if (d <= fabs(a.r - b.r)) {        double r = a.r < b.r ? a.r : b.r;        return PI * r * r;    }    double ang1 = acos((a.r * a.r + d * d - b.r * b.r) / 2. / a.r / d);    double ang2 = acos((b.r * b.r + d * d - a.r * a.r) / 2. / b.r / d);    double ret = ang1 * a.r * a.r + ang2 * b.r * b.r - d * a.r * sin(ang1);    return ret;}bool check(Round& ori) {    repf (i, 0, n - 1) {        if (insection_of_rounds(ori, r[i]) * 2 < PI * r[i].r * r[i].r)            return false;    }    return true;};double bin(double l, double r, Round& ori) {    double mid;    while (fabs(l - r) >= EPS) {        mid = (l + r) / 2;        ori.r = mid;        if (check(ori)) {            r = mid;        } else {            l = mid + EPS;        }    }    return mid;}int main() {    scanf("%d", &t);    while (t--) {        scanf("%d", &n);        repf (i, 0, n - 1) {            scanf("%lf%lf%lf", &r[i].x, &r[i].y, &r[i].r);        }        double ans = 1e10;        repf (i, 0, n - 1) {            ori.x = r[i].x;            ori.y = r[i].y;            double right = 0;            repf (j, 0, n - 1) {                right = max(right, dis(ori, r[j]) + r[j].r);            }            ans = min(ans, bin(0, right, ori));        }        printf("%.4f\n", ans);    }    return 0;}