UVA - 143 Orchard Trees （点在三角形内）

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：

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 using namespace std;
 typedef long long ll;
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const int maxp = ;
 int sgn(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }
 struct Point{
     double x, y;
     Point(){}
     Point(double _x, double _y){
         x = _x, y = _y;
     }
     void input(){
         scanf("%lf%lf", &x, &y);
     }
     bool operator == (Point b) const{
         return sgn(x - b.x) ==  && sgn(y - b.y) == ;
     }
     bool operator < (Point b)const{
         return sgn(x - b.x) ==  ? sgn(y - b.y < ) : x < b.x;
     }
     Point operator - (const Point &b)const{
         return Point(x - b.x, y - b.y);
     }
     //²æ»ý
     double operator ^(const Point &b){
         return x * b.y - y * b.x;
     }
     //µã»ý
     double operator *(const Point &b){
         return x * b.x + y * b.y;
     }
     double len(){
         return hypot(x, y);
     }
     double len2(){
     return x * x + y * y;
     }
     double distant(Point p){
         return hypot(x - p.x, y - p.y);
     }
     Point operator + (const Point &b)const{
         return Point (x + b.x, y + b.y);
     }
     Point operator * (const double &k)const{
         return Point(x * k, y * k);
     }
     Point operator / (const double &k)const{
         return Point(x / k, y / k);
     }
 };
 struct Line{
     Point s, e;
     Line(){}
     Line(Point _s, Point _e){s = _s, e = _e;}
     bool operator == (Line v){
         return (s == v.s) && (e == v.e);
     }
     bool pointonseg(Point p){
         return sgn((p - s)^(e - s)) ==  && sgn((p - e)*(p - s)) <= ;
     }

 };
 struct polygon{
     int n;
     Point p[maxp];
     Line l[maxp];
     void add(Point q){
         p[n ++] = q;
     }
     void input(int _n){
         n = _n;
         for(int i = ;i < n;i++) p[i].input();
     }
      void getline(){
         for(int i = ;i < n;i++){
             l[i] = Line(p[i], p[(i+) % n]);
         }
     }
     int relationpoint(Point q){
         for(int i = ;i < n;i++){
             if(p[i] == q) return ;
         }
         getline();
         for(int i = ;i < n;i++){
             if(l[i].pointonseg(q)) return ;
         }
         int cnt = ;
         for(int i = ;i < n;i++){
             int j = (i + ) % n;
             int k = sgn((q - p[j])^(p[i] - p[j]));
             int u = sgn(p[i].y - q.y);
             int v = sgn(p[j].y - q.y);
             if(k >  && u <  && v >= ) cnt++;
             if(k <  && v <  && u >= ) cnt--;
         }
         return cnt != ;
     }
 };
 int main()
 {
     double x1, x2, x3, y1, y2, y3;
     polygon a;
     while(~scanf("%lf%lf%lf%lf%lf%lf",&x1, &y1, &x2, &y2, &x3, &y3) && (x1|| x2|| x3|| y1|| y2|| y3))
     {
         a.n = ;
         a.add(Point(x1,y1));
         a.add(Point(x2,y2));
         a.add(Point(x3,y3));
         int cnt = ;
         for(double i = ;i <= ;i++)
             for(double j = ;j <= ;j++)
                 if(a.relationpoint(Point(i,j))) cnt ++;
         printf("%4d\n",cnt);
     }
     return ;
 }