UESTC 33 Area --凸包面积

题意： 求一条直线分凸包两边的面积。

解法： 因为题意会说一定穿过，那么不会有直线与某条边重合的情况。我们只要找到一个直线分成的凸包即可，另一个的面积等于总面积减去那个的面积。

怎么得到分成的一个凸包呢？

从0~n扫过去，如果扫到的边与直线不相交，那么把端点加进新凸包中，如果直线与扫到的边相交了，那么就将交点加入新凸包，然后以后不相交的话也不加入点到新凸包中，直到遇到下一个与直线相交的边，则把交点又加入新凸包，然后在扫到末尾加入点。这样就得到了。

即找到如图：

注意四舍五入。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define eps 1e-8
using namespace std;

struct Point{
    double x,y;
    Point(double x=, double y=):x(x),y(y) {}
    void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
int dcmp(double x) {
    if(x < -eps) return -;
    if(x > eps) return ;
    return ;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) ==  && dcmp(a.y-b.y) == ; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

Point DisP(Point A,Point B) { return Length(B-A); }
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
    return max(A.x,B.x) >= min(C.x,D.x) &&
           max(C.x,D.x) >= min(A.x,B.x) &&
           max(A.y,B.y) >= min(C.y,D.y) &&
           max(C.y,D.y) >= min(A.y,B.y) &&
           dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <=  &&
           dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= ;
}
void SegIntersectionPoint(Point& P,Point a,Point b,Point c,Point d) {  //需保证ab,cd相交
    P.x = (Cross(d-a,b-a)*c.x - Cross(c-a,b-a)*d.x)/(Cross(d-a,b-a)-Cross(c-a,b-a));
    P.y = (Cross(d-a,b-a)*c.y - Cross(c-a,b-a)*d.y)/(Cross(d-a,b-a)-Cross(c-a,b-a));
}
double CalcConvexArea(Point* p,int n)
{
    double area = 0.0;
    for(int i=;i<n-;i++)
        area += Cross(p[i]-p[],p[i+]-p[]);
    return fabs(area*0.5);
}
Point p[],ch[];
Point P,A,B;

int main()
{
    int n,i,m;
    while(scanf("%d",&n)!=EOF && n)
    {
        for(i=;i<n;i++) p[i].input();
        A.input(), B.input();
        Point tmpA = B+(A-B)*, tmpB = A+(B-A)*;
        A = tmpA, B = tmpB;
        double Total = CalcConvexArea(p,n);
        int tot = , fir = , add = ;
        ch[tot++] = p[];
        for(i=;i<n;i++)
        {
            Point C = p[i], D = p[(i+)%n];
            if(SegmentIntersection(A,B,C,D))
            {
                SegIntersectionPoint(P,A,B,C,D);
                ch[tot++] = P;
                if(!fir) fir = ;
                else     fir = , add = ;
                if(P == D) i++;
            }
            else if(!fir) ch[tot++] = p[(i+)%n];
            if(add) ch[tot++] = p[(i+)%n];
        }
        double Now = CalcConvexArea(ch,tot);
        double Other = Total-Now;
        int N = (int)(Now+0.5), O = (int)(Other+0.5);
        if(O > N) swap(N,O);
        printf("%d %d\n",N,O);
    }
    return ;
}