hdu 4491 Windmill Animation（几何+模拟）

给定平面上的n(n<=20)个点,不存在三点共线。然后每次用某条直线（原点及斜率）逆时针旋转，每碰到一个点后，更换原点跟斜率。求出前s个原点。

考虑到n很小，可以模拟来搞。每次得到一个原点及斜率，然后构造出在直线上原点(O)的上方(rp）和下方(rn)两个点。然后依次每个不在该直线上的点，如果该点在直线上方，那么要达到该点逆时针旋转的角度为向量 rp->O 和向量 p->O的夹角。反之则是向量O->rp和向量O->rn的夹角。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<fstream>
#include<sstream>
#include<cstdlib>
#include<vector>
#include<string>
#include<cstdio>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define MP make_pair
using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);
int dcmp(double x) {
  if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

struct Point {
  double x, y;
  Point(double x=0, double y=0):x(x),y(y) { }
};

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); }

bool operator < (const Point& a, const Point& b) {
  return a.x < b.x || (a.x == b.x && a.y < b.y);
}

bool operator == (const Point& a, const Point &b) {
  return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

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 Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double angle(Vector v) { return atan2(v.y, v.x); }
double torad(double ang)
{
    return ang / 180 * PI;
}

void getLineGeneralEquation(const Point& p1, const Point& p2, double& a, double& b, double &c) {
  a = p2.y-p1.y;
  b = p1.x-p2.x;
  c = -a*p1.x - b*p1.y;
}

Point read_point()
{
    Point a;
    scanf("%lf%lf", &a.x, &a.y);
    return a;
}

const int maxn = 30;
int P, cas, ind, n, s, st, ans[maxn];
double ang;
Point p[maxn];

int main()
{
    scanf("%d", &P);
    while(P--)
    {
        scanf("%d%d%d%d%lf", &cas, &n, &s, &st, &ang);
        ang = torad(ang);
        FF(i, 1, n+1)
        {
            scanf("%d", &ind);
            p[ind] = read_point();
        }
        int tot = 0, id1 = st, id2 = st;
        Point now = p[st];
        while(tot < s)
        {
            int nxt;
            double a, b, c, tmp, tang, nxtang = 1e50;
            Point O = now, rp = O + Point(1000*cos(ang), 1000*sin(ang)) , rn = O - Point(1000*cos(ang), 1000*sin(ang));
            getLineGeneralEquation(O, rp, a, b, c);
            FF(i, 1, n+1) if(i != id1 && i != id2)
            {
                tmp = a*p[i].x + b*p[i].y + c;
                if(tmp > 0) tang = Angle(rn-O, p[i]-O);
                else if(tmp < 0) tang = Angle(rp-O, p[i]-O);
                else tang = 0;
                if(dcmp(nxtang - tang) > 0) nxtang = tang, nxt = i;
            }
            ans[tot++] = nxt;
            ang = angle(now-p[nxt]);
            now = p[nxt];
            id1 = id2;
            id2 = nxt;
        }
        printf("%d", cas);
        REP(i, tot) printf(" %d", ans[i]);
        puts("");
    }
    return 0;
}