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=0, double y=0):x(x),y(y) {} void input() { scanf("%lf%lf",&x,&y); } }; typedef Point Vector; int dcmp(double x) { if(x < -eps) return -1; if(x > eps) return 1; return 0; } 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) == 0 && dcmp(a.y-b.y) == 0; } 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)) <= 0 && dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0; } 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=1;i<n-1;i++) area += Cross(p[i]-p[0],p[i+1]-p[0]); return fabs(area*0.5); } Point p[25],ch[25]; Point P,A,B; int main() { int n,i,m; while(scanf("%d",&n)!=EOF && n) { for(i=0;i<n;i++) p[i].input(); A.input(), B.input(); Point tmpA = B+(A-B)*20003, tmpB = A+(B-A)*20003; A = tmpA, B = tmpB; double Total = CalcConvexArea(p,n); int tot = 0, fir = 0, add = 0; ch[tot++] = p[0]; for(i=0;i<n;i++) { Point C = p[i], D = p[(i+1)%n]; if(SegmentIntersection(A,B,C,D)) { SegIntersectionPoint(P,A,B,C,D); ch[tot++] = P; if(!fir) fir = 1; else fir = 0, add = 1; if(P == D) i++; } else if(!fir) ch[tot++] = p[(i+1)%n]; if(add) ch[tot++] = p[(i+1)%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 0; }
时间: 2024-10-16 00:23:44