题目大意：有两块木板交叉起来接雨水，问最多能接多少。

 

分析：题目描述很简单，不过有些细节还是需要注意到，如下图几种情况：

 

#include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int MAXN = 107;const double oo = 1e4+7;const double EPS = 1e-8;int sign(double val){    if(val > EPS)return 1;    if(fabs(val) < EPS)return 0;    return -1;}struct Point{    double x, y;    Point(double x=0, double y=0):x(x), y(y){}    Point operator - (const Point &tmp) const{        return Point(x-tmp.x, y-tmp.y);    }    double operator ^(const Point &tmp) const{        return x*tmp.y - y*tmp.x;    }    bool operator == (const Point &tmp) const{        return fabs(x-tmp.x) < EPS && fabs(y-tmp.y) < EPS;    }    double operator *(const Point &tmp) const{        return x*tmp.x + y*tmp.y;    }};struct Segment{    Point s, e;    double a, b, c;///ax + by = c    Segment(Point s=0, Point e=0):s(s), e(e){        a = s.y - e.y;        b = e.x - s.x;        c = e.x*s.y - s.x*e.y;    }    bool Inter(const Segment &t)const{        int v1 = sign((s-e)^(t.s-e));        int v2 = sign((s-e)^(t.e-e));        if(!v1 && !v2)return false;///共线        if(!v1 && t.s.x >= min(s.x, e.x) && t.s.x <= max(s.x, e.x)               && t.s.y >= min(s.y, e.y) && t.s.y <= max(s.y, e.y)        || !v2 && t.e.x >= min(s.x, e.x) && t.e.x <= max(s.x, e.x)               && t.e.y >= min(s.y, e.y) && t.e.y <= max(s.y, e.y)        || v1 * v2 == -1)return true;        return false;    }    Point CrossNode(const Segment &t) const{        Point ans;        ans.x = (c*t.b-t.c*b)/(a*t.b-t.a*b);        ans.y = (c*t.a-t.c*a)/(b*t.a-t.b*a);        return ans;    }};double Dist(Point a, Point b){    return sqrt((a-b) * (a-b));}double Find(Point crs, Point p[], int N){    double sum = 0;    for(int i=0; i
          
           =min(p[i].x, p[j].x) && crs.x<=max(p[i].x, p[j].x)        || crs.x>=p[i].x && crs.x>=p[j].x && (k>0&&(p[i].x-p[j].x>EPS) || k<0&&(p[j].x-p[i].x>EPS))        || crs.x<=p[i].x && crs.x<=p[j].x && (k<0&&(p[j].x-p[i].x>EPS) || k>0&&(p[i].x-p[j].x>EPS)))        if(p[i].y-crs.y > EPS && p[j].y-crs.y > EPS)        {            Point A = p[i].y < p[j].y ? p[i] : p[j];            Point B = (A == p[i] ? p[j] : p[i]);            Segment t1(Point(-oo, A.y), Point(oo, A.y));            Segment t2(crs, B);            B = t1.CrossNode(t2);            double La = Dist(A, B);            double Lb = Dist(A, crs);            double Lc = Dist(B, crs);            double p = (La+Lb+Lc) / 2;            sum += sqrt(p*(p-La)*(p-Lb)*(p-Lc));        }    }    return sum;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        Point p[MAXN], crs;        scanf("%lf%lf%lf%lf", &p[0].x, &p[0].y, &p[1].x, &p[1].y);        scanf("%lf%lf%lf%lf", &p[2].x, &p[2].y, &p[3].x, &p[3].y);        Segment L1(p[0], p[1]), L2(p[2], p[3]);        double ans = 0;        if(L1.Inter(L2) && L2.Inter(L1))        {            crs = L1.CrossNode(L2);            ans = Find(crs, p, 4);        }        printf("%.2f\n", ans+EPS);    }    return 0;}