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 = 4010;
const double eps = 5*1e-13;
const double PI = acos(-1.0);

double 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;

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)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(const Point& p, const Point& a1, const Point& a2)
{
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

// 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 = 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(n > 1) m--;
    ch.resize(m);
    return ch;
}

int IsPointInPolygon(const Point& p, const vector<Point>& poly)
{
    int wn = 0;
    int n = poly.size();
    for(int i = 0; i < n; i++)
    {
        const Point& p1 = poly[i];
        const Point& p2 = poly[(i+1)%n];
        if(p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1; // 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 > 0 && d1 <= 0 && d2 > 0) wn++;
        if(k < 0 && d2 <= 0 && d1 > 0) wn--;
    }
    if (wn != 0) return 1; // 内部
    return 0; // 外部
}

bool ConvexPolygonDisjoint(const vector<Point> ch1, const vector<Point> ch2)
{
    int c1 = ch1.size();
    int c2 = ch2.size();
    for(int i = 0; i < c1; i++)
        if(IsPointInPolygon(ch1[i], ch2) != 0) return false; // 内部或边界上
    for(int i = 0; i < c2; i++)
        if(IsPointInPolygon(ch2[i], ch1) != 0) return false; // 内部或边界上
    for(int i = 0; i < c1; i++)
        for(int j = 0; j < c2; j++)
            if(SegmentProperIntersection(ch1[i], ch1[(i+1)%c1], ch2[j], ch2[(j+1)%c2])) return false;
    return true;
}

int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m) == 2 && n > 0 && m > 0)
    {
        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(ConvexPolygonDisjoint(ConvexHull(P1), ConvexHull(P2)))
            printf("Yes
");
        else
            printf("No
");
    }
    return 0;
}