UVA 10256 The Great Divide (判断凸包相交)

Posted 2022-08-12 wulitaotao

题目链接：UVA 10256

题意

有n个红点和m个蓝点,问是否存在一条直线能够分开红点和蓝点。

题解

分别求出红点和蓝点的凸包，判断两个凸包是否相交。

凸包不相交的条件：

• 凸包上的任意点都在另一个凸包的外面
• 凸包的任意线段都与另一个凸包不相交

代码

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const double pi = acos(-1.0);
class Point 
public:
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) 
    Point operator+(Point a) 
        return Point(a.x + x, a.y + y);
    
    Point operator-(Point a) 
        return Point(x - a.x, y - a.y);
    
    bool operator<(const Point &a) const 
        if (x == a.x)
            return y < a.y;
        return x < a.x;
    
    bool operator==(const Point &a) const 
        if (fabs(x - a.x) < eps && fabs(y - a.y) < eps)
            return 1;
        return 0;
    
    double length() 
        return sqrt(x * x + y * y);
    
;

typedef Point Vector;

double cross(Vector a, Vector b) 
    return a.x * b.y - a.y * b.x;


double dot(Vector a, Vector b) 
    return a.x * b.x + a.y * b.y;


bool isclock(Point p0, Point p1, Point p2) 
    Vector a = p1 - p0;
    Vector b = p2 - p0;
    if (cross(a, b) < -eps)
        return true;
    return false;


double getDistance(Point a, Point b) 
    return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));


typedef vector<Point> Polygon;
Polygon Andrew(Polygon s) 
    Polygon u, l;
    if(s.size() < 3) return s;
    sort(s.begin(), s.end());
    u.push_back(s[0]);
    u.push_back(s[1]);
    l.push_back(s[s.size() - 1]);
    l.push_back(s[s.size() - 2]);
    for(int i = 2 ; i < s.size() ; ++i) 
        for(int n = u.size() ; n >= 2 && !isclock(u[n - 2], u[n - 1], s[i]); --n) 
            u.pop_back();
        
        u.push_back(s[i]);
    
    for(int i = s.size() - 3 ; i >= 0 ; --i) 
        for(int n = l.size() ; n >=2 && !isclock(l[n-2],l[n-1],s[i]); --n) 
            l.pop_back();
        
        l.push_back(s[i]);
    
    for(int i = 1 ; i < u.size() - 1 ; i++) l.push_back(u[i]);
    return l;


int dcmp(double x)  
    if (fabs(x) <= eps)
        return 0;
    return x > 0 ? 1 : -1;


// 判断点在线段上
bool OnSegment(Point p, Point a1, Point a2) 
    return dcmp(cross(a1 - p, a2 - p)) == 0 && dcmp(dot(a1 - p, a2 - p)) < 0;


// 判断线段相交
bool Intersection(Point a1, Point a2, Point b1, 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;


// 判断点在凸包内
int isPointInPolygon(Point p, vector<Point> s) 
    int wn = 0, cc = s.size();
    for (int i = 0; i < cc; i++) 
        Point p1 = s[i];
        Point p2 = s[(i + 1) % cc];
        if (p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1;
        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;
    return 0;


void solve(Polygon s1, Polygon s2) 
    int c1 = s1.size(), c2 = s2.size();
    for(int i = 0; i < c1; ++i) 
        if(isPointInPolygon(s1[i], s2)) 
            printf("No\n");
            return;
        
    
    for(int i = 0; i < c2; ++i) 
        if(isPointInPolygon(s2[i], s1)) 
            printf("No\n");
            return;
        
    
    for (int i = 0; i < c1; i++) 
        for (int j = 0; j < c2; j++) 
            if (Intersection(s1[i], s1[(i + 1) % c1], s2[j], s2[(j + 1) % c2])) 
                printf("No\n");
                return;
            
        
    
    printf("Yes\n");


int main() 
    int n, m;
    while (cin >> n >> m) 
        if(n == 0 && m == 0) break;
        Polygon s1, s2;
        for (int i = 0; i < n; ++i) 
            double x1, x2;
            scanf("%lf%lf", &x1, &x2);
            s1.push_back(Point(x1, x2));
        
        for (int i = 0; i < m; ++i) 
            double x1, x2;
            scanf("%lf%lf", &x1, &x2);
            s2.push_back(Point(x1, x2));
        
        if(s1.size()) s1 = Andrew(s1);
        if(s2.size()) s2 = Andrew(s2);
        solve(s1, s2);
    
    return 0;

参考

《算法竞赛入门经典》 刘汝佳 / 陈锋