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UVa 10256 (判断两个凸包相离) The Great Divide

题意:

给出n个红点,m个蓝点。问是否存在一条直线使得红点和蓝点分别分布在直线的两侧,这些点不能再直线上。

分析:

求出两种点的凸包,如果两个凸包相离的话,则存在这样一条直线。

判断凸包相离需要判断这两件事情:

  1. 任何一个凸包的任何一个顶点不能在另一个凸包的内部或者边界上。
  2. 两个凸包的任意两边不能相交。

二者缺一不可,第一条很好理解,但为什么还要判断第二条,因为存在这种情况:

虽然每个凸包的顶点都在另一个凸包的外部,但两个凸包明显是相交的。

 

  1 //#define LOCAL  2 #include <cstdio>  3 #include <cstring>  4 #include <algorithm>  5 #include <cmath>  6 #include <vector>  7 using namespace std;  8   9 const int maxn = 500 + 10; 10 const double eps = 1e-10; 11 const double PI = acos(-1.0); 12  13 int dcmp(double x) 14 { 15     if(fabs(x) < eps)    return 0; 16     else return x < 0 ? -1 : 1; 17 } 18  19 struct Point 20 { 21     double x, y; 22     Point(double x=0, double y=0):x(x), y(y) {} 23 }; 24 typedef Point Vector; 25 Point operator + (Point a, Point b) { return Point(a.x+b.x, a.y+b.y); } 26 Point operator - (Point a, Point b) { return Point(a.x-b.x, a.y-b.y); } 27 Point operator * (Point a, double p) { return Point(a.x*p, a.y*p); } 28 Point operator / (Point a, double p) { return Point(a.x/p, a.y/p); } 29 bool operator < (Point a, Point b) { return a.x < b.x || (a.x == b.x && a.y < b.y); } 30 bool operator == (Point a, Point b) { return a.x == b.x && a.y == b.y; } 31  32 double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } 33 double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } 34  35 bool SegmentIntersection(Point a1, Point a2, Point b1, Point b2) 36 { 37     double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1); 38     double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1); 39     return dcmp(c1*c2) < 0 && dcmp(c3*c4) < 0; 40 } 41  42 bool OnSegment(Point p, Point a, Point b) 43 {//点是否在线段上,不包含在端点处的情况  44     return dcmp(Cross(a-p, b-p)) == 0 && dcmp(Dot(a-p, b-p)) < 0; 45 } 46  47 vector<Point> ConvexHull(vector<Point> p) 48 { 49     sort(p.begin(), p.end()); 50     p.erase(unique(p.begin(), p.end()), p.end()); 51      52     int n = p.size(); 53     int m = 0; 54     vector<Point> ch(n+1); 55     for(int i = 0; i < n; ++i) 56     { 57         while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; 58         ch[m++] = p[i]; 59     } 60     int k = m; 61     for(int i = n-2; i >= 0; --i) 62     { 63         while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; 64         ch[m++] = p[i]; 65     } 66     if(m > 1) m--; 67     ch.resize(m); 68     return ch; 69 } 70  71 int IsPointInPolygon(Point p, const vector<Point>& poly) 72 { 73     int wn = 0; 74     int n = poly.size(); 75     for(int i = 0; i < n; ++i) 76     { 77         if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1;    //边界  78         int k = dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i])); 79         int d1 = dcmp(poly[i].y-p.y); 80         int d2 = dcmp(poly[(i+1)%n].y-p.y); 81         if(k > 0 && d1 <= 0 && d2 > 0) wn++; 82         if(k < 0 && d2 <= 0 && d1 > 0) wn--; 83     } 84     if(wn)    return 1;    //内部  85     return 0;            //外部  86 } 87  88 bool ConvexPolygonDisjiont(const vector<Point> ch1, const vector<Point> ch2) 89 { 90     int c1 = ch1.size(), c2 = ch2.size(); 91     for(int i = 0; i < c1; ++i) 92         if(IsPointInPolygon(ch1[i], ch2) != 0) return false; 93     for(int i = 0; i < c2; ++i) 94         if(IsPointInPolygon(ch2[i], ch1) != 0) return false; 95     for(int i = 0; i < c1; ++i) 96           for(int j = 0; j < c2; ++j) 97             if(SegmentIntersection(ch1[i], ch1[(i+1)%c1], ch2[j], ch2[(j+1)%c2])) return false; 98     return true; 99 }100 101 int main(void)102 {103     #ifdef LOCAL104         freopen("10256in.txt", "r", stdin);105     #endif106     107     int n, m;108     while(scanf("%d%d", &n, &m) == 2)109     {110         if(!n && !m) break;111         vector<Point> p1, p2;112         double x, y;113         for(int i = 0; i < n; ++i)114         {115             scanf("%lf%lf", &x, &y);116             p1.push_back(Point(x, y));117         }118         for(int i = 0; i < m; ++i)119         {120             scanf("%lf%lf", &x, &y);121             p2.push_back(Point(x, y));122         }123         if(ConvexPolygonDisjiont(ConvexHull(p1), ConvexHull(p2)))    puts("Yes");124         else puts("No");125     }126     127     return 0;128 }
代码君

 

UVa 10256 (判断两个凸包相离) The Great Divide