【poj3831】Open-air shopping malls

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给一些圆，求一个大圆使得对每个圆都覆盖了至少一半，求其最小半径，大圆的圆心要是一个小圆的圆心。

直接枚举圆心，二分半径，相交区域是两个拱形。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#define Rep(i, x, y) for (int i = x; i <= y; i ++)#define Dwn(i, x, y) for (int i = x; i >= y; i --)#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)#define eps 0.000001using namespace std;typedef long long LL;typedef double DB;const int N = 25;int T, n; DB R[N], X[N], Y[N], ans, PI = 3.1415926535;DB angle(DB a, DB b, DB c) { return acos((a*a + b*b - c*c) / (2*a*b)); }DB sq(DB x) { return x * x; }DB Dist(int x, int y) { return sqrt(sq(X[x] - X[y]) + sq(Y[x] - Y[y])); }bool Check(int x, DB Rx) {Rep(i, 1, n) {DB d = Dist(i, x), an1, an2, s1, s2, s0;if (Rx >= R[i] + d) continue ;if (Rx <= d) return 0;if (Rx + d <= R[i] + eps) {if (Rx * Rx * 2 < R[i] * R[i]) return 0;continue ;}s0 = R[i] * R[i] * PI;an1 = angle(d, R[i], Rx), an2 = angle(d, Rx, R[i]);s1 = R[i] * R[i] * an1 - R[i] * R[i] * sin(2*an1) / 2;s2 = Rx * Rx * an2 - Rx * Rx * sin(2*an2) / 2;if (2 * (s1 + s2) < s0) return 0;}return 1;}int main(){scanf ("%d", &T);while (T --) {scanf ("%d", &n);Rep(i, 1, n) {scanf ("%lf%lf%lf", &X[i], &Y[i], &R[i]);}ans = 12000;Rep(i, 1, n) {DB l = 0, r = 12000;while (l + eps < r) {DB mid = (l + r) / 2;if (Check(i, mid)) r = mid;else l = mid;}ans = min(ans, l);}printf("%.4f\n", ans);}return 0;}

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