这是一道较短的编程问题，请问有人可以解答一下吗？谢谢。用C++

Question

完整题目都在图片里，谢谢。

Answer 1

题目完整了么？下面没有explanation之类的了？目测就是判断点对和多边形的边有没有交点

#include <stdio.h>
#include <iostream>
using namespace std;

struct point  
{  
    double x,y;  
};  
struct segment  
{  
    point begin,end;  
};  
double min(double x,double y)  
{  
    return x<y?x:y;  
}  
double max(double x,double y)  
{  
    return x>y?x:y;  
}  
bool onsegment(point pi,point pj,point pk) //判断点pk是否在线段pi pj上   
{  
    if(min(pi.x,pj.x)<=pk.x&&pk.x<=max(pi.x,pj.x))  
    {  
        if(min(pi.y,pj.y)<=pk.y&&pk.y<=max(pi.y,pj.y))  
        {  
            return true;  
        }  
    }  
    return false;  
}  
double direction(point pi,point pj,point pk) //计算向量pkpi和向量pjpi的叉积   
{  
    return (pi.x-pk.x)*(pi.y-pj.y)-(pi.y-pk.y)*(pi.x-pj.x);  
}  
bool judge(point p1,point p2,point p3,point p4) //判断线段p1p2和p3p4是否相交   
{  
    double d1 = direction(p3,p4,p1);  
    double d2 = direction(p3,p4,p2);  
    double d3 = direction(p1,p2,p3);  
    double d4 = direction(p1,p2,p4);  
    if(d1*d2<0&&d3*d4<0)  
        return true;  
    if(d1==0&&onsegment(p3,p4,p1))  
        return true;  
    if(d2==0&&onsegment(p3,p4,p2))  
        return true;  
    if(d3==0&&onsegment(p1,p2,p3))  
        return true;  
    if(d4==0&&onsegment(p1,p2,p4))  
        return true;  
    return false;  
} 

int main()
{
    int n, m;
    cin >> n >> m;
    point * v = new point[n];
    point * p = new point[m];
    for (int i = 0; i < n; i++)
    {
        int x, y;
        cin >> x >> y;
        v[i].x = x;
        v[i].y = y;
    }
    for (int i = 0; i < m; i++)
    {
        int x, y;
        cin >> x >> y;
        p[i].x = x;
        p[i].y = y;
    }
    int cnt = 0;
    for (int a = 0; a < m - 1; a++)
        for (int b = a + 1; b < m; b++)
        {
            int flag = 0;
            for (int c = 0; c < n - 1; c++)
            {
                if (judge(p[a], p[b], v[c], v[c + 1]))
                { flag = 1; break; }
            }
            if (!flag)
                if (judge(p[a], p[b], v[0], v[n-1])) flag = 1;
            if (!flag) cnt++;
        }
    cout << cnt << endl;
    return 0;
}