UVA - 143 Orchard Trees （点在三角形内）

Posted 2022-08-23 carered

 

 

题意：

　给出三角形的三个点的坐标（浮点数），
    问落在三角形内及三角形边上的整点有多少？

思路：所有点暴力判断（点的范围1-99，三角形可能是0-100，因为这个WA了一下orz）

AC代码：

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<algorithm>
  4 #include<iostream>
  5 #include<cstring>
  6 using namespace std;
  7 typedef long long ll;
  8 const double eps = 1e-8;
  9 const double pi = acos(-1.0);
 10 const int maxp = 5000;
 11 int sgn(double x)
 12 
 13     if(fabs(x) < eps) return 0;
 14     else return x < 0 ? -1 : 1;
 15 
 16 struct Point
 17     double x, y;
 18     Point()
 19     Point(double _x, double _y)
 20         x = _x, y = _y;
 21     
 22     void input()
 23         scanf("%lf%lf", &x, &y);
 24     
 25     bool operator == (Point b) const
 26         return sgn(x - b.x) == 0 && sgn(y - b.y) == 0;
 27     
 28     bool operator < (Point b)const
 29         return sgn(x - b.x) == 0 ? sgn(y - b.y < 0) : x < b.x;
 30     
 31     Point operator - (const Point &b)const
 32         return Point(x - b.x, y - b.y);
 33     
 34     //²æ»ý
 35     double operator ^(const Point &b)
 36         return x * b.y - y * b.x;
 37     
 38     //µã»ý
 39     double operator *(const Point &b)
 40         return x * b.x + y * b.y;
 41     
 42     double len()
 43         return hypot(x, y);
 44     
 45     double len2()
 46     return x * x + y * y;
 47     
 48     double distant(Point p)
 49         return hypot(x - p.x, y - p.y);
 50     
 51     Point operator + (const Point &b)const
 52         return Point (x + b.x, y + b.y);
 53     
 54     Point operator * (const double &k)const
 55         return Point(x * k, y * k);
 56     
 57     Point operator / (const double &k)const
 58         return Point(x / k, y / k);
 59     
 60 ;
 61 struct Line
 62     Point s, e;
 63     Line()
 64     Line(Point _s, Point _e)s = _s, e = _e;
 65     bool operator == (Line v)
 66         return (s == v.s) && (e == v.e);
 67     
 68     bool pointonseg(Point p)
 69         return sgn((p - s)^(e - s)) == 0 && sgn((p - e)*(p - s)) <= 0;
 70     
 71 
 72 ;
 73 struct polygon
 74     int n;
 75     Point p[maxp];
 76     Line l[maxp];
 77     void add(Point q)
 78         p[n ++] = q;
 79     
 80     void input(int _n)
 81         n = _n;
 82         for(int i = 0;i < n;i++) p[i].input();
 83     
 84      void getline()
 85         for(int i = 0;i < n;i++)
 86             l[i] = Line(p[i], p[(i+1) % n]);
 87         
 88     
 89     int relationpoint(Point q)
 90         for(int i = 0;i < n;i++)
 91             if(p[i] == q) return 3;
 92         
 93         getline();
 94         for(int i = 0;i < n;i++)
 95             if(l[i].pointonseg(q)) return 2;
 96         
 97         int cnt = 0;
 98         for(int i = 0;i < n;i++)
 99             int j = (i + 1) % n;
100             int k = sgn((q - p[j])^(p[i] - p[j]));
101             int u = sgn(p[i].y - q.y);
102             int v = sgn(p[j].y - q.y);
103             if(k > 0 && u < 0 && v >= 0) cnt++;
104             if(k < 0 && v < 0 && u >= 0) cnt--;
105         
106         return cnt != 0;
107     
108 ;
109 int main()
110 
111     double x1, x2, x3, y1, y2, y3;
112     polygon a;
113     while(~scanf("%lf%lf%lf%lf%lf%lf",&x1, &y1, &x2, &y2, &x3, &y3) && (x1|| x2|| x3|| y1|| y2|| y3))
114     
115         a.n = 0;
116         a.add(Point(x1,y1));
117         a.add(Point(x2,y2));
118         a.add(Point(x3,y3));
119         int cnt = 0;
120         for(double i = 1;i <= 99;i++)
121             for(double j = 1;j <= 99;j++)
122                 if(a.relationpoint(Point(i,j))) cnt ++;
123         printf("%4d\\n",cnt);
124     
125     return 0;
126