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  • POJ 3130 How I Mathematician Wonder What You Are! 半平面交

    和POJ3335一样,只不过这题是逆时针

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #define eps 1e-18
    using namespace std;
    
    const int MAXN = 105;
    double a, b, c;
    int n, cnt;
    
    struct Point
    {
        double x, y;
    }point[MAXN], p[MAXN], tp[MAXN];
    
    void Get_equation(Point p1, Point p2)
    {
        a = p2.y - p1.y;
        b = p1.x - p2.x;
        c = p2.x * p1.y - p1.x * p2.y;
    }
    
    Point Intersection(Point p1, Point p2)
    {
        double u = fabs(a * p1.x + b * p1.y + c);
        double v = fabs(a * p2.x + b * p2.y + c);
        Point t;
        t.x = (p1.x * v + p2.x * u) / (u + v);
        t.y = (p1.y * v + p2.y * u) / (u + v);
        return t;
    }
    
    void Cut()
    {
        int tmp = 0;
        for(int i=1; i<=cnt; i++)
        {
            //顺时针是>-eps和>eps,逆时针是<eps和<-eps
            if(a * p[i].x + b * p[i].y + c < eps) tp[++tmp] = p[i];
            else
            {
                if(a * p[i-1].x + b * p[i-1].y + c < -eps)
                    tp[++tmp] = Intersection(p[i-1], p[i]);
                if(a * p[i+1].x + b * p[i+1].y + c < -eps)
                    tp[++tmp] = Intersection(p[i], p[i+1]);
            }
        }
        for(int i=1; i<=tmp; i++)
            p[i] = tp[i];
        p[0] = p[tmp];
        p[tmp+1] = p[1];
        cnt = tmp;
    }
    
    void solve()
    {
        for(int i=1; i<=n; i++)
            p[i] = point[i];
        point[n+1] = point[1];
        p[0] = p[n];
        p[n+1] = p[1];
        cnt = n;
        for(int i=1; i<=n; i++)
        {
            Get_equation(point[i], point[i+1]);
            Cut();
        }
    }
    
    int main()
    {
        while(~scanf("%d", &n) && n)
        {
            for(int i=1; i<=n; i++)
                scanf("%lf%lf", &point[i].x, &point[i].y);
            solve();
            puts(cnt > 0? "1" : "0");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pach/p/7459307.html
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