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。

//#define LOCAL
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;

const int maxn = 500 + 10;
const double eps = 1e-10;
const double PI = acos(-1.0);

int dcmp(double x)
{
    if(fabs(x) < eps)    return 0;
    else return x < 0 ? -1 : 1;
}

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x), y(y) {}
};
typedef Point Vector;
Point operator + (Point a, Point b) { return Point(a.x+b.x, a.y+b.y); }
Point operator - (Point a, Point b) { return Point(a.x-b.x, a.y-b.y); }
Point operator * (Point a, double p) { return Point(a.x*p, a.y*p); }
Point operator / (Point a, double p) { return Point(a.x/p, a.y/p); }
bool operator < (Point a, Point b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator == (Point a, Point b) { return a.x == b.x && a.y == b.y; }

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

bool SegmentIntersection(Point a1, Point a2, Point b1, Point b2)
{
    double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);
    double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
    return dcmp(c1*c2) < 0 && dcmp(c3*c4) < 0;
}

bool OnSegment(Point p, Point a, Point b)
{//点是否在线段上，不包含在端点处的情况 
    return dcmp(Cross(a-p, b-p)) == 0 && dcmp(Dot(a-p, b-p)) < 0;
}

vector<Point> ConvexHull(vector<Point> p)
{
    sort(p.begin(), p.end());
    p.erase(unique(p.begin(), p.end()), p.end());
    
    int n = p.size();
    int m = 0;
    vector<Point> ch(n+1);
    for(int i = 0; i < n; ++i)
    {
        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
    for(int i = n-2; i >= 0; --i)
    {
        while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    if(m > 1) m--;
    ch.resize(m);
    return ch;
}

int IsPointInPolygon(Point p, const vector<Point>& poly)
{
    int wn = 0;
    int n = poly.size();
    for(int i = 0; i < n; ++i)
    {
        if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1;    //边界 
        int k = dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i]));
        int d1 = dcmp(poly[i].y-p.y);
        int d2 = dcmp(poly[(i+1)%n].y-p.y);
        if(k > 0 && d1 <= 0 && d2 > 0) wn++;
        if(k < 0 && d2 <= 0 && d1 > 0) wn--;
    }
    if(wn)    return 1;    //内部 
    return 0;            //外部 
}

bool ConvexPolygonDisjiont(const vector<Point> ch1, const vector<Point> ch2)
{
    int c1 = ch1.size(), c2 = ch2.size();
    for(int i = 0; i < c1; ++i)
        if(IsPointInPolygon(ch1[i], ch2) != 0) return false;
    for(int i = 0; i < c2; ++i)
        if(IsPointInPolygon(ch2[i], ch1) != 0) return false;
    for(int i = 0; i < c1; ++i)
          for(int j = 0; j < c2; ++j)
            if(SegmentIntersection(ch1[i], ch1[(i+1)%c1], ch2[j], ch2[(j+1)%c2])) return false;
    return true;
}

int main(void)
{
    #ifdef LOCAL
        freopen("10256in.txt", "r", stdin);
    #endif
    
    int n, m;
    while(scanf("%d%d", &n, &m) == 2)
    {
        if(!n && !m) break;
        vector<Point> p1, p2;
        double x, y;
        for(int i = 0; i < n; ++i)
        {
            scanf("%lf%lf", &x, &y);
            p1.push_back(Point(x, y));
        }
        for(int i = 0; i < m; ++i)
        {
            scanf("%lf%lf", &x, &y);
            p2.push_back(Point(x, y));
        }
        if(ConvexPolygonDisjiont(ConvexHull(p1), ConvexHull(p2)))    puts("Yes");
        else puts("No");
    }
    
    return 0;
}