HDU 4998 Rotate

题目链接

思路：构造一个多边形，顶点集由边界顶点和旋转中心点组成。将此多边形绕每个中心点旋转相应角度，得到新的多边形，任取原多边形和新多边形的两条对应边，分别作中垂线，直到两中垂线平行为止。两中垂线的交点即为旋转中心A，旋转角p为每个旋转角的和，注意将其对2π取模，p = fmod(p, 2*acos(-1.0));

//Rotate.cpp#include <ios>#include <map>#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cerrno>#include <cfloat>#include <cstdio>#include <cwchar>#include <iosfwd>#include <limits>#include <locale>#include <memory>#include <string>#include <vector>#include <cassert>#include <ciso646>#include <climits>#include <clocale>#include <complex>#include <csetjmp>#include <csignal>#include <cstdarg>#include <cstddef>#include <cstdlib>#include <cstring>#include <cwctype>#include <fstream>#include <iomanip>#include <istream>#include <numeric>#include <ostream>#include <sstream>#include <utility>#include <iostream>#include <iterator>#include <valarray>#include <algorithm>#include <exception>#include <stdexcept>#include <streambuf>#include <functional>#define LL long long#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1#define PI 3.1415926535897932626#define mid(n) ((center[n]+pol.a[n])*0.5)#define CLEAR(name, init) memset(name, init, sizeof(name))const double eps = 1e-8;const int MAXN = (int)1e9 + 5;using namespace std;struct Point {    double x, y;    Point(double x = 0.0, double y = 0.0): x(x), y(y) { }};struct Polygon {    int n;    Point a[20];    Polygon() {}};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }Vector operator - (Point A, Point B) { return Vector(A.x-B.x, A.y-B.y); }Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }double Length(Vector A) { return sqrt(Dot(A, A)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }Point rotate(Vector A, double alpha) {    double cos_rad = cos(alpha), sin_rad = sin(alpha);    return Vector(A.x*cos_rad-A.y*sin_rad, A.x*sin_rad+A.y*cos_rad);}Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {    Vector u = P - Q;    double t = Cross(w, u) / Cross(v, w);    return P + v*t;}struct line {    Point a,b;    line(){}    line(Point x,Point y):a(x),b(y){}};int dblcmp(double x) {    if (fabs(x) < eps) return 0;    return x > 0 ? 1 : -1;}bool parallel(line a,line b) {    return !dblcmp(Cross(a.a-a.b,b.a-b.b));}int main(int argc, char const *argv[]) {#ifndef ONLINE_JUDGE    freopen("D:\\Documents\\Programs\\Acm\\input.txt", "r", stdin);#endif    int T; cin >> T;    while(T--) {        int n; cin >> n;        Point left_up = Point(0.0, 100.0);        Point right_up = Point(100.0, 100.0);        Point left_down = Point(0.0, 0.0);        Point right_down = Point(100.0, 0.0);        Polygon pol;        pol.n = n + 4;        pol.a[0] = left_up; pol.a[1] = right_up;        pol.a[2] = left_down; pol.a[3] = right_down;        Point center[20];        double rad[20], p = 0.0;        for(int i = 0; i < n; i++) {            scanf("%lf%lf%lf", ¢er[i+4].x, ¢er[i+4].y, &rad[i+4]);            pol.a[i+4] = center[i+4];            p += rad[i+4];        }        for(int i = 0; i < 4; i++) center[i] = pol.a[i];        while(p >= 2*acos(-1.0)) p -= 2*acos(-1.0);        for(int i = 4; i < pol.n; i++) {            for(int j = 0; j < pol.n; j++) {                pol.a[j] = rotate(pol.a[j]-center[i], rad[i]) + center[i];            }        }        Point res;        for (int i = 0; i < pol.n; i++) {            for (int j = i+1; j < pol.n; j++) {                Point p1 = (center[i]+pol.a[i]) * 0.5;                Point q1 = (center[j]+pol.a[j]) * 0.5;                Point p2 = rotate(center[i]-p1, PI/2.0)+p1;                Point q2 = rotate(center[j]-q1, PI/2.0)+q1;                line P = line(p1, p2), Q = line(q1, q2);                if (parallel(P,Q)) continue;                double s1=Cross(p1-q1, q2-q1);                double s2=Cross(p2-q1, q2-q1);                res = (p2*s1-p1*s2)/(s1-s2);            }        }        printf("%.10lf %.10lf %.10lf\n", res.x, res.y, p);    }    return 0;}