题目大意就是两根木块组成一个槽，问槽里能装多少雨水,注意雨水垂直落下，思路也很简单，就是分类讨论有点糟。
1.如果两条线段不相交或者平行，则装0；
2.有一条平行x轴，装0；
3.若上面覆盖下面的，装0；
4.其它，叉积求面积。
直接上代码：
#include
#include
#include
using namespace std; const double eps=1e-8; struct point{ double x; double y; }; struct line{ point a; point b; }l1,l2; double ans; //求叉积 double xmult(point p0 ,point p1 ,point p2){ return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } int dblcmp(double n){ if(fabs(n)
0?1:-1; } //判断是否相交，如何相交 int judge(line l1 ,line l2){ double d1=dblcmp(max(l1.a.x,l1.b.x)-min(l2.a.x,l2.b.x)); double d2=dblcmp(max(l2.a.x,l2.b.x)-min(l1.a.x,l1.b.x)); double d3=dblcmp(max(l1.a.y,l1.b.y)-min(l2.a.y,l2.b.y)); double d4=dblcmp(max(l2.a.y,l2.b.y)-min(l1.a.y,l1.b.y)); double d5=dblcmp(xmult(l2.a,l1.a,l1.b)); double d6=dblcmp(xmult(l2.b,l1.a,l1.b)); double d7=dblcmp(xmult(l1.a,l2.a,l2.b)); double d8=dblcmp(xmult(l1.b,l2.a,l2.b)); if(d1>=0&&d2>=0&&d3>=0&&d4>=0){ if(d5*d6>0||d7*d8>0) return 0;//不相交 else if(d5==0&&d6==0) return 1;//共线相交 else if(d5==0||d6==0||d7==0||d8==0) return 2;//端点相交 else return 3;//规范相交 } return 0; } //求斜率 bool getslope(line l ,double &k){ double t=l.a.x-l.b.x; if(t==0) return false; k=(l.a.y-l.b.y)/t; return true; } //求线段交点 point getIntersect(line l1, line l2) { point p; double A1 = l1.b.y - l1.a.y; double B1 = l1.a.x - l1.b.x; double C1 = (l1.b.x - l1.a.x) * l1.a.y - (l1.b.y - l1.a.y) * l1.a.x; double A2 = l2.b.y - l2.a.y; double B2 = l2.a.x - l2.b.x; double C2 = (l2.b.x - l2.a.x) * l2.a.y - (l2.b.y - l2.a.y) * l2.a.x; p.x = (C2*B1 - C1*B2) / (A1*B2 - A2*B1); p.y = (C1*A2 - C2*A1) / (A1*B2 - A2*B1); if(p.x==-0) p.x=0; return p; } //求a,b两点中y坐标更大的点 point getbiggerY(point a ,point b){ point q; if (dblcmp(a.y-b.y) > 0) { q.x = a.x; q.y = a.y; } else { q.x = b.x; q.y = b.y; } return q; } double getarea(point p ,point p1 ,point p2){ point q; double a; if(dblcmp(p1.y-p2.y)>=0) { q.y = p2.y; q.x = p.x+(p1.x-p.x)*(p2.y-p.y) / (p1.y-p.y); //求另一点的坐标 a=fabs(xmult(p, p2, q)) / 2; //叉积求面积 } else { q.y = p1.y; q.x = p.x+(p2.x-p.x)*(p1.y-p.y) / (p2.y-p.y); a=fabs(xmult(p, p1, q)) / 2; } return a; } int main() { int t; cin>>t; while(t--){ point p1 ,p2 ,p; cin>>l1.a.x>>l1.a.y>>l1.b.x>>l1.b.y; cin>>l2.a.x>>l2.a.y>>l2.b.x>>l2.b.y; if(judge(l1,l2)<=1)//是否相交 {ans=0;cout<<0<
0) { //当两条线段的斜率符号相同时， int d1 = dblcmp(k1-k2); int d2 = dblcmp(k2); if (d1>0&&d2>0&&dblcmp(p2.x-p1.x)*dblcmp(p2.x-p.x)<=0 || d1<0&&d2>0&&dblcmp(p1.x-p2.x)*dblcmp(p1.x-p.x)<= 0 || d1>0&&d2<0&&dblcmp(p1.x-p2.x)*dblcmp(p1.x-p.x)<= 0 || d1<0&&d2<0&&dblcmp(p2.x-p1.x)*dblcmp(p2.x-p.x)<= 0)//覆盖情况 // { ans = 0;//cout<<3<