AOJ0187 Stoning Fortune

解法

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どれか二線の交点がなかったら凶。
あったら、3つの交点を求めてヘロンの公式S=√s(s-a)(s-b)(s-c)で面積を求めてはっぴー

ソース

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int main(){
  L ls[3];
  int x1[3], y1[3], x2[3], y2[3];
  while( cin >> x1[0] >> y1[0] >> x2[0] >> y2[0], x1[0]|y1[0]|x2[0]|y2[0] ){
    for(int i = 1 ; i < 3 ; i++ ){
      cin >> x1[i] >> y1[i] >> x2[i] >> y2[i];
    }
    for(int i = 0 ; i < 3 ; i++ ){
      ls[i] = L( P( x1[i], y1[i]), P( x2[i], y2[i]));
    }
 
    G triangle(3);
    bool flag = true;
    for(int i = 0 ; i < 3 ; i++ ){
      const int NEXT = ( i + 1 ) % 3;
      if(!intersect( ls[i], ls[NEXT])){
        flag = false;
        break;
      }else{
        triangle[i] = crosspoint( ls[i], ls[NEXT]);
      }
    }
 
    double s = (abs(triangle[0]-triangle[1])+abs(triangle[1]-triangle[2])+abs(triangle[2]-triangle[0])) / 2;
    double area = sqrt(s * ( s - abs(triangle[0] - triangle[1])) * ( s - abs(triangle[1] - triangle[2])) * ( s - abs(triangle[2] - triangle[0])));
    if(!flag || area < EPS) cout << "kyo" << endl;
    else if(area < 100000) cout << "syo-kichi" << endl;
    else if(area < 1000000) cout << "kichi" << endl;
    else if(area < 1900000) cout << "chu-kichi" << endl;
    else cout << "dai-kichi" << endl;
  }
}