UVA - 12301 - An Angular Puzzle （计算几何~平面三角）

标签：uva   acm   计算几何   

题目地址：点这里

思路：可以先确定A，B的坐标，然后再通过确定向量来硬算出角度。。好像可以推公式做，没推出来╮(╯_╰)╭

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;

const double PI = 4 * atan(1.0);

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 - (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 == b.x && a.y < b.y);
} 

const double eps = 1e-10;
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; }
double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }

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 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;
} 
 
double DistanceToLine(Point P, Point A, Point B) {  
    Vector v1 = B-A, v2 = P - A;  
    return fabs(Cross(v1,v2) / Length(v1)); 
}  

double DistanceToSegment(Point P, Point A, Point B) {  
    if(A==B) return Length(P-A);  
    Vector v1 = B - A, v2 = P - A, v3 = P - B;  
    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);  
    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);  
    else return fabs(Cross(v1, v2)) / Length(v1);  
}  

Point GetLineProjection(Point P, Point A, Point B) {
	Vector v = B - A;
	return A + v * ( Dot(v, P-A) / Dot(v, v) ); 
}  

bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
	double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
			c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
	return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
} 

bool OnSegment(Point p, Point a1, Point a2) {
	return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
} 

double ConvexPolygonArea(Point* p, int n) {  
    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;  
}  


int main() {
	double a, b, c, d, e;
	while(scanf("%lf %lf %lf %lf %lf", &a, &b, &c, &d, &e) != EOF && a) {
		if(dcmp(a + b + c + d + e - 180) != 0) {
			printf("Impossible\n"); continue;
		}
		a = a / 180 * PI;
		b = b / 180 * PI;
		c = c / 180 * PI;
		d = d / 180 * PI;
		e = e / 180 * PI;
		Point A = Point(0,0);  
        Point B = Point(1,0);  
        Vector AC = Vector(1,tan(b+c));  
        Vector BC = Vector(1,-tan(d+e));  
        Vector AE = Vector(1,tan(c));  
        Vector BD = Vector(1,-tan(e));  
          
        Point E = GetLineIntersection(A, AE, B, BC);
        Point D = GetLineIntersection(A, AC, B, BD);
		  
        double ans = Angle(D-E, A-E);  
        printf("%.2lf\n", ans / PI * 180);  
	}
	return 0;
} 

UVA - 12301 - An Angular Puzzle （计算几何~平面三角）

标签：uva   acm   计算几何   

原文地址：http://blog.csdn.net/u014355480/article/details/43715255