poj1279

题意：计算多边形核的面积。

分析：半平面交的模板。有两个问题要注意，1.题目没说多边形点的顺序是顺时针还是逆时针，要先用面积的正负来判断点的顺序。2.题目中说坐标都在16位整数范围内，也就是说半平面交模板中初始的无限大平面的四个顶点设为±1e5就可以了，原本设为±1e15无限WA。

#include <cstdio>
#include <algorithm>
#define vector point
using std::swap;
struct point
{
    double x,y;
    point(double xx = 0,double yy = 0)
    {
        x = xx;
        y = yy;
    }
    point operator - (const point& s)
    {
        return point(x - s.x, y - s.y);
    }
    point operator + (const point& s)
    {
        return point(x + s.x,y + s.y);
    }
};
struct polygon
{
    point p[2000];
    int size;
};
struct line
{
    point first,second;
    line(point p1 = point(),point p2 = point())
    {
        first = p1;
        second = p2;
    }
};

double cross_product(vector v1,vector v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

//求两直线交点
point line_intersection(line ln1,line ln2)
{
    double a1,b1,c1,a2,b2,c2;
    a1 = ln1.first.y - ln1.second.y;
    b1 = ln1.second.x - ln1.first.x;
    c1 = ln1.first.x * ln1.second.y - ln1.second.x * ln1.first.y;
    a2 = ln2.first.y - ln2.second.y;
    b2 = ln2.second.x - ln2.first.x;
    c2 = ln2.first.x * ln2.second.y - ln2.second.x * ln2.first.y;
    double d = a1 * b2 - a2 * b1;
    return point((b1 * c2 - b2 * c1) / d,(c1 * a2 - c2 * a1) / d);
}

//一个多边形与一个半平面的交集
polygon hpi(polygon& poly,line& ln)
{
    int m = 0;
    polygon hull;
    point p1 = ln.first,p2 = ln.second;
    //穷举多边形的所有点，判断是否在半平面上
    //如果凸多边形hull与直线ln有交点就求交点
    for(int i = 0;i < poly.size;i++)
    {
        double c = cross_product(p2 - p1,poly.p[i] - p1);
        double d = cross_product(p2 - p1,poly.p[(i + 1) % poly.size] - p1);
        //点poly.p[i]在半平面上
        if(c >= 0)
            hull.p[m++] = poly.p[i];
        //有交点
        if(c * d < 0)
            hull.p[m++] = line_intersection(line(poly.p[(i + 1) % poly.size],poly.p[i]),ln);
    }
    hull.size = m;
    return hull;
}

//poly的顶点顺时针
bool polygon_kernel(polygon& poly,polygon& knl)
{
    //初始化核为无限大
    knl.p[0] = point(-1e5,-1e5);
    knl.p[1] = point(1e5,-1e5);
    knl.p[2] = point(1e5,1e5);
    knl.p[3] = point(-1e5,1e5);
    knl.size = 4;
    line ln;
    point pre = poly.p[0];
    for(int i = 1;i <= poly.size;i++)
    {
        ln.first = poly.p[i % poly.size];
        ln.second = poly.p[i - 1];
        knl = hpi(knl,ln);
        if(knl.size == 0)
            return false;
    }
    return true;
}

double polygon_area(polygon& poly)
{
    double s = 0;
    for(int i = 0;i < poly.size - 1;i++)
        s += cross_product(poly.p[i],poly.p[i + 1]);
    s += cross_product(poly.p[poly.size - 1],poly.p[0]);
    return s / 2;
}

int main()
{
    int t,n;
    polygon gallery,kernel;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 0;i < n;i++)
            scanf("%lf%lf",&gallery.p[i].x,&gallery.p[i].y);
        gallery.size = n;
        if(!polygon_kernel(gallery,kernel))
        {
            int left = 0,right = n - 1;
            while(left < right)
                swap(gallery.p[left++],gallery.p[right--]);
            polygon_kernel(gallery,kernel);
        }
        double s;
        s = polygon_area(kernel);
        printf("%.2lf\n",s);
    }
    return 0;
}