poj 1106 Transmitters (枚举+叉积运用)
题目链接:http://poj.org/problem?id=1106
算法思路:由于圆心和半径都确定,又是180度,这里枚举过一点的直径,求出这个直径的一个在圆上的端点,就可以用叉积的大于,等于,小于0判断点在直径上,左,右。 这里要记录直径两边的加直径上的点的个数,去最大的。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; const double eps = 1e-8; const double PI = acos(-1.0); const double INF = 1000000000000000.000; struct Point{ double x,y; Point(double x=0, double y=0) : x(x),y(y){ } //构造函数 }; typedef Point Vector; struct Circle{ Point c; double r; Circle() {} Circle(Point c,double r): c(c),r(r) {} }; Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);} Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);} Vector operator * (double p,Vector A){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); } int dcmp(double x){ if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } bool operator == (const Point& a, const Point& b){ return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; } ///向量(x,y)的极角用atan2(y,x); inline double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; } inline double Length(Vector A) { return sqrt(Dot(A,A)); } inline double Angle(Vector A, Vector B) { return acos(Dot(A,B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y - A.y * B.x; } Vector vecunit(Vector v){ return v / Length(v);} //单位向量 Point read_point(){ Point A; scanf("%lf %lf",&A.x,&A.y); return A; } //多边形 //求面积 double PolygonArea(Point* p,int n){ //n个点 double area = 0; for(int i=1;i<n-1;i++){ area += Cross(p[i]-p[0],p[i+1]-p[0]); } return area/2; } /*************************************分 割 线*****************************************/ int main() { //freopen("E:\\acm\\input.txt","r",stdin); Point O,P[155]; double R; while(scanf("%lf %lf %lf",&O.x,&O.y,&R) == 3 && dcmp(R)>0) { int N; cin>>N; int cnt = 0; for(int i=1;i<=N;i++) { Point temp = read_point(); double len = Length(temp-O); if(dcmp(len-R) > 0) continue; P[++cnt] = temp; } int ans = 0; for(int i=1; i<=cnt; i++) { if(P[i] == O) { ans = max(ans,1); continue; } int lnum = 1; int rnum = 1; Vector v = P[i] - O; v = (-1.0)*vecunit(v); Point T = O + R*v; for(int j=1; j<=cnt; j++) { if(i == j) continue; if(dcmp(Cross(P[j]-T,O-T)) > 0 ) lnum++; else if(dcmp(Cross(P[j]-T,O-T)) == 0) lnum++,rnum++; else rnum++; } int num = max(lnum,rnum); ans = max(ans,num); } printf("%d\n",ans); } }