uva 10256 The Great Divide

题意：给定两个点集，一个红点集，另一个蓝点集，询问，能否找到一条直线能，使得任取一个红点和蓝点都在直线异侧。

思路：划分成两个凸包，一个红包，一个蓝包。两个凸包不相交不重合。

1.任取一个凸包中的点不在另一个凸包中。

2.任取一个凸包中的边与另一个凸包不相交。

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 #define up(i,x,y) for(i=x;i<=y;i++)
 #define w(a) while(a)
 using namespace std;
 const double inf=0x3f3f3f3f;
 const int N = ;
 const double eps = *1e-;
 const double PI = acos(-1.0);

 double dcmp(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }

 struct Point
 {
     double x, y;
     Point(double x=, double y=):x(x),y(y) { }
 };

 typedef Point Vector;

 Vector operator - (const Point& A, const Point& B)
 {
     return Vector(A.x-B.x, A.y-B.y);
 }

 double Cross(const Vector& A, const Vector& B)
 {
     return A.x*B.y - A.y*B.x;
 }

 double Dot(const Vector& A, const Vector& B)
 {
     return A.x*B.x + A.y*B.y;
 }

 bool operator < (const Point& p1, const Point& p2)
 {
     return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
 }

 bool operator == (const Point& p1, const Point& p2)
 {
     return p1.x == p2.x && p1.y == p2.y;
 }

 bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const 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)< && dcmp(c3)*dcmp(c4)<;
 }

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

 // 点集凸包
 // 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
 // 如果不介意点集被修改，可以改成传递引用
 vector<Point> ConvexHull(vector<Point> p)
 {
     // 预处理，删除重复点
     sort(p.begin(), p.end());
     p.erase(unique(p.begin(), p.end()), p.end());

     int n = p.size();
     int m = ;
     vector<Point> ch(n+);
     for(int i = ; i < n; i++)
     {
         while(m >  && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     int k = m;
     for(int i = n-; i >= ; i--)
     {
         while(m > k && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     if(n > ) m--;
     ch.resize(m);
     return ch;
 }

 int IsPointInPolygon(const Point& p, const vector<Point>& poly)
 {
     int wn = ;
     int n = poly.size();
     for(int i = ; i < n; i++)
     {
         const Point& p1 = poly[i];
         const Point& p2 = poly[(i+)%n];
         if(p1 == p || p2 == p || OnSegment(p, p1, p2)) return -; // 在边界上
         int k = dcmp(Cross(p2-p1, p-p1));
         int d1 = dcmp(p1.y - p.y);
         int d2 = dcmp(p2.y - p.y);
         if(k >  && d1 <=  && d2 > ) wn++;
         if(k <  && d2 <=  && d1 > ) wn--;
     }
     if (wn != ) return ; // 内部
     return ; // 外部
 }

 bool ConvexPolygonDisjoint(const vector<Point> ch1, const vector<Point> ch2)
 {
     int c1 = ch1.size();
     int c2 = ch2.size();
     for(int i = ; i < c1; i++)
         if(IsPointInPolygon(ch1[i], ch2) != ) return false; // 内部或边界上
     for(int i = ; i < c2; i++)
         if(IsPointInPolygon(ch2[i], ch1) != ) return false; // 内部或边界上
     for(int i = ; i < c1; i++)
         for(int j = ; j < c2; j++)
             if(SegmentProperIntersection(ch1[i], ch1[(i+)%c1], ch2[j], ch2[(j+)%c2])) return false;
     return true;
 }

 int main()
 {
     int n, m;
     while(scanf("%d%d", &n, &m) ==  && n >  && m > )
     {
         vector<Point> P1, P2;
         double x, y;
         for(int i = ; i < n; i++)
         {
             scanf("%lf%lf", &x, &y);
             P1.push_back(Point(x, y));
         }
         for(int i = ; i < m; i++)
         {
             scanf("%lf%lf", &x, &y);
             P2.push_back(Point(x, y));
         }
         if(ConvexPolygonDisjoint(ConvexHull(P1), ConvexHull(P2)))
             printf("Yes\n");
         else
             printf("No\n");
     }
     return ;
 }