An Easy Problem?!(poj2826线段相交,直线线段相交)
题意：给你两个同样长的木板，围城一个多边形，水从空中垂直下落，问木板能盛放多少体积的水。
思路：分类讨论
1.如果线段没有交点，那么面积就是0
2.如果线段有交点，但是有一个模板被另一个木板覆盖还是接不到水，怎么判断了，过两个木板的最高的两个点做垂线，如果垂线和和交点到另一木板的顶点这段相交，说明木板有覆盖，不行可以画个图试试，但是覆盖不一定遮挡，所以要判断交点是在另一个木板顶点的上下，在上面说明木板被覆盖，面积为0，反之没有
3.因为盛水是水平的，所以过另个木板的顶点做水平线，如果水平线的交点在交点和顶点之间，那么面积就是这三个顶点的面积
这题需要直线与线段相交，线段与线段相交，点在线段和点在直线上。
直线与线段相交，只要有一个叉乘小于0即可，但是有可能叉乘等于0，一个点在直线上，还要判断点是否在线段上，用点到直线的距离，距离为0就在，反之就不再
点到直线距离，面积除以底。
直线与线段的交点，求出直线与直线的交点，判断他是否在线段上，判线段上还要注意端点，
线段与线段的交点，也是求出直线与直线的交点 ，判断他是否同时在两个线段上，或者可以先判断两个线段是否相交，然后直接求交点
点在线段上，叉积和点积同时用并且在判断是否在两端点
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;struct Point{ double x,y; Point(double x = 0,double y = 0):x(x),y(y){}};typedef Point Vector;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) ;}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 ;}const double eps = 1e-8;int dcmp(double x){ if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;}bool operator == (Point a,Point b){ return dcmp(a.x-b.x) == 0&& dcmp(a.y-b.y) == 0;}bool operator < (Point a,Point b){ return a.x < b.x || (a.x == b.x && a.y < b.y);}bool Onsegment(Point p,Point a,Point b){ return dcmp(Cross(b-a,p-a)) == 0 && dcmp(Dot(b-p,a-p)) < 0 || (p == a) || (p == b);}bool OnLine(Point p,Point a,Point b){ return fabs(Cross(p-a,a-b)) / Length(b-a);}bool SegmentIntersection(Point a,Point b,Point c,Point d){ double d1 = Cross(b-a,c-a),d2 = Cross(b-a,d-a),d3 = Cross(d-c,a-c),d4 = Cross(d-c,b-c); if(dcmp(d1)*dcmp(d2) < 0 || dcmp(d3)*dcmp(d4) < 0) return true; else if(dcmp(d1) == 0 && !OnLine(c,a,b) ) return true; else if(dcmp(d2) == 0 && !OnLine(d,a,b) ) return true; else if(dcmp(d3) == 0 && !OnLine(a,c,d) ) return true; else if(dcmp(d4) == 0 && !OnLine(b,c,d) ) return true; else return false;}bool Segmentsection(Point a,Point b,Point c,Point d){ double d1 = Cross(b-a,c-a),d2 = Cross(b-a,d-a),d3 = Cross(d-c,a-c),d4 = Cross(d-c,b-c); if(dcmp(d1)*dcmp(d2) < 0 && dcmp(d3)*dcmp(d4) < 0) return true; else if(dcmp(d1) == 0 && Onsegment(c,a,b) ) return true; else if(dcmp(d2) == 0 && Onsegment(d,a,b) ) return true; else if(dcmp(d3) == 0 && Onsegment(a,c,d) ) return true; else if(dcmp(d4) == 0 && Onsegment(b,c,d) ) return true; else return false;}bool pointInpoly(Point p,Point a,Point b,Point c,Point d){ double d1 = Cross(b-a,p-a); double d2 = Cross(d-c,p-c); double d3 = Cross(c-b,p-b); double d4 = Cross(a-d,p-d); return dcmp(d1)*dcmp(d2) > 0 && dcmp(d3)*dcmp(d4) > 0;}Point Segment(Point p,Vector v,Point q,Vector w){ Vector u = p-q; double t = Cross(w,u) / Cross(v,w); return p + v*t;}struct Line{ Point s,e; Line(Point s = 0,Point e = 0) :s(s),e(e){}};Point Max(Point a,Point b){ return a.y > b.y ? a : b;}int main(){ int n; scanf("%d",&n); while(n--) { Point a,b,c,d; scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y); scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y); Point P1 = Max(a,b); Point P2 = Max(c,d); if(!Segmentsection(a,b,c,d)) printf("0.00\n"); else if(a.y == b.y || c.y == d.y) printf("0.00\n"); else if(!Cross(b-a,c-d)) printf("0.00\n"); else { Point k = Segment(a,b-a,c,d-c); Point P3 = Point(P1.x,P1.y+1); Point P4 = Point(P2.x,P2.y+1); if(SegmentIntersection(P1,P3,c,d)) { Point k1 = Segment(P1,P3-P1,c,d-c); if(Onsegment(k1,P2,k)) { if(dcmp(P1.y-k1.y) <= 0) { printf("0.00\n");continue; } } } if(SegmentIntersection(P2,P4,a,b)) { Point k1 = Segment(P2,P4-P2,a,b-a); if(Onsegment(k1,P1,k)) { if(dcmp(P2.y-k1.y) <= 0) { printf("0.00\n");continue; } } } P3 = Point(P1.x+1,P1.y); P4 = Point(P2.x+1,P2.y); if(SegmentIntersection(P1,P3,c,d)) { Point k1 = Segment(P1,P3-P1,c,d-c); if(Onsegment(k1,P2,k)) printf("%.2lf\n",fabs(Cross(P1-k,k1-k))/2);continue; } if(SegmentIntersection(P2,P4,a,b)) { Point k1 = Segment(P2,P4-P2,a,b-a); if(Onsegment(k1,P1,k)) printf("%.2lf\n",fabs(Cross(P2-k,k1-k))/2);continue; } } } return 0;}