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  • SGU 120 Archipelago (简单几何)

    120. Archipelago

    time limit per test: 0.25 sec. 
    memory limit per test: 4096 KB

    Archipelago Ber-Islands consists of N islands that are vertices of equiangular and equilateral N-gon. Islands are clockwise numerated. Coordinates of island N1 are (x1, y1), and island N2 – (x2, y2). Your task is to find coordinates of all N islands.

    Input

    In the first line of input there are N, N1 and N2 (3£ N£ 150, 1£ N1,N2£N, N1¹N2separated by spaces. On the next two lines of input there are coordinates of island N1 and N2 (one pair per line) with accuracy 4digits after decimal point. Each coordinate is more than -2000000 and less than 2000000.

    Output

    Write N lines with coordinates for every island. Write coordinates in order of island numeration. Write answer with 6 digits after decimal point.

    Sample Input

    4 1 3
    1.0000 0.0000
    1.0000 2.0000
    

    Sample Output

    1.000000 0.000000
    0.000000 1.000000
    1.000000 2.000000
    2.000000 1.000000
    
     



    题意:给你正N边形上两个点,按顺时针给出。让你按顺时针输出这N个点
    
    
    思路:用向量N1N2的中垂线 和 向量N1N2旋转(n2-n1)*PI/n的交点就可以求出圆心。求出圆心后用向量n1O旋转N遍就可以。


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #define REP(_,a,b) for(int _ = (a); _ <= (b); _++)
    using namespace std;
    const double eps = 1e-10;
    const int maxn = 160;
    const double PI = acos(-1.0);
    double ang,rad;
    int n,n1,n2;
    struct Point{
    	double x,y;
    	Point(double x=0.0,double y = 0.0):x(x),y(y){}
    }P[maxn];
    typedef Point Vector;
    
    struct Line {
    	Point P;
    	Vector v;
    	double ang;
    	Line(){}
    	Line(Point P,Vector v):P(P),v(v){
    		ang = atan2(v.y,v.x);
    	}
    	bool operator <(const Line&L) const{
    		return ang < L.ang;
    	}
    };
    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 * (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==a.y && 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;
    }
    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 Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}
    Vector Rotate(Vector A,double rad) {return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); }
    Vector Normal(Vector A) {
    	double L = Length(A);
    	return Vector(-A.y/L,A.x/L);
    }
    Point GetIntersection(Line a,Line b){
    	Vector u = a.P-b.P;
    	double t = Cross(b.v,u) / Cross(a.v,b.v);
    	return a.P+a.v*t;
    }
    int main(){
    	while(~scanf("%d%d%d",&n,&n1,&n2)){
    		ang = (n2-n1)*PI/n;
    		rad = 2*PI/n;
    		scanf("%lf%lf%lf%lf",&P[n1].x,&P[n1].y,&P[n2].x,&P[n2].y);
    		Line a = Line((P[n1]+P[n2])/2,Normal(P[n2]-P[n1]));
    		Line b = Line(P[n1],Rotate(Normal(P[n2]-P[n1]),ang));
    		Point o = GetIntersection(a,b);
    		Vector t = P[n1]-o;
    		int d = n1+1,cnt = 1;
    		while(d != n1){
    			P[d] = o+Rotate(t,-cnt*rad);
    			d = d%n + 1;
    			cnt++;
    		}
    		REP(i,1,n) {
    			printf("%.6lf %.6lf
    ",P[i].x,P[i].y);
    		}
    	}
    	return 0;
    }
    


     


    题意:给你正N边形上两个点。按顺时针给出,让你按顺时针输出这N个点
    
    
    思路:用向量N1N2的中垂线 和 向量N1N2旋转(n2-n1)*PI/n的交点就可以求出圆心。求出圆心后用向量n1O旋转N遍就可以。

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  • 原文地址:https://www.cnblogs.com/liguangsunls/p/6694230.html
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