AOJ Problem 0187 : Stoning Fortune

問題文:404 Not Found

ライブラリゲー。ccwやintersectSS,crosspointはspaghetti sourceのもの。

本選でのこれの配点があんまり高くないのは悲しいよね。

気をつけるべきは「0,0,0,1」とかの入力を見逃さないこと。

#include <iostream>
#include <vector>
#include <complex>
using namespace std;
static const double EPS = 1e-12;
#define rep(i,n) for(int i=0;i<n;i++)

const double INF = 1e12;
typedef complex<double> P,point;
namespace std {bool operator < (const P& a, const P& b) {return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);}}
double cross(const P& a, const P& b) {return imag(conj(a)*b);}
double dot(const P& a, const P& b) {return real(conj(a)*b);}
struct L : public vector<P> {L(const P &a, const P &b) {push_back(a); push_back(b);}};

int ccw(P a, P b, P c) {
  b -= a; c -= a;
  if (cross(b, c) > 0)   return +1;       // counter clockwise
  if (cross(b, c) < 0)   return -1;       // clockwise
  if (dot(b, c) < 0)     return +2;       // c--a--b on line
  if (norm(b) < norm(c)) return -2;       // a--b--c on line
  return 0;
}

bool intersectSS(const L &s, const L &t) {
	return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 && ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}
P crosspoint(const L &l, const L &m) {
  double A = cross(l[1] - l[0], m[1] - m[0]);
  double B = cross(l[1] - l[0], l[1] - m[0]);
  if (abs(A) < EPS && abs(B) < EPS) return m[0]; // same line
  return m[0] + B / A * (m[1] - m[0]);
}
double getArea(P a, P b , P c){
		b-=a, c-=a;
		return fabs( b.real() * c.imag() - c.real() * b.imag() ) / 2.0; 
}
int main(){
	double a,b,c,d;
	while(cin >> a >> b >> c >> d , a || b || c || d){
		L one = L(P(a,b),P(c,d));
		cin >> a >> b >> c >> d;
		L two = L(P(a,b),P(c,d));
		cin >> a >> b >> c >> d;
		L three = L(P(a,b),P(c,d));
		double ret = 0;
		
		if(intersectSS(one,two) && intersectSS(two,three) && intersectSS(one,three)){
			ret = getArea( crosspoint(one,two) , crosspoint(two,three) , crosspoint(one,three) );
		}
		if(ret >= 1900000){
			cout << "dai-kichi" << endl;
		}else if(ret >= 1000000){
			cout << "chu-kichi" << endl;
		}else if(ret >= 100000){
			cout << "kichi" << endl;
		}else if(ret > 0){
			cout << "syo-kichi" << endl;
		}else{
			cout << "kyo" << endl;
		}
	}
}