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：
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)<eps) return 0;return 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)-min(l2));double d2=dblcmp(max(l2)-min(l1));double d3=dblcmp(max(l1)-min(l2));double d4=dblcmp(max(l2)-min(l1));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=lif(t==0) return false;k=(l)/t;return true;}//求线段交点point getIntersect(line l1, line l2) {point p;double A1 = l1double B1 = l1double C1 = (l1) * l1- l1double A2 = l2double B2 = l2double C2 = (l2) * l2- l2p.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>>l1cin>>l2>>l2if(judge(l1,l2)<=1)//是否相交ans=0；else{p1=getbiggerY(l1.a ,l1.b);p2=getbiggerY(l2.a ,l2.b);p=getIntersect(l1 ,l2); if(!dblcmp(p.x-p1.x)&&!dblcmp(p.y-p1.y) || !dblcmp(p.x-p2.x)&&!dblcmp(p.x-p2.y))//开口方向ans=0;else{double k1 ,k2;bool f1 ,f2;f1 = getslope(l1, k1);f2 = getslope(l2, k2);if (!dblcmp(k1) || !dblcmp(k2)) ans=0;else if (f1 && f2) {//如果两条线都不与y轴平行if (dblcmp(k1*k2) > 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;elseans=getarea(p,p1,p2);}elseans=getarea(p,p1,p2);}elseans=getarea(p,p1,p2);}}printf (“%.2lf\n”, ans);}return 0;}
有了你，我不再作孤飞于蓝天的雄鹰，
