Poj2826 An Easy Problem

  呵呵哒。WA了无数次,一开始想的办法最终发现都有缺陷。首先需要知道:

    1)线段不相交,一定面积为0

    2)有一条线段与X轴平行,面积一定为0

    3)线段相交,但是能接水的三角形上面线段把下面的线段完全覆盖。

(1),(2)的情况简单,主要是解决(3)。下面对(3)进行讨论,如下图所示,设p1,p2是两线段各自位置较高的点,p0为两线段的交点,向量e是三角形p0p1p2的关于边p1p2的外侧法向量。则当e的终点位于第1,2象限时才会有积水,3,4象限是没有的。判断e的方向可以根据它与(p0p1+p0p2)的点积,e与(p0p1+p0p2)的点积总是大于0的。

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const double PI = acos(-1.0);
const int N = 300;
const double EPS = 1e-8;//实数精度
//点结构类型
struct Point{
    double x, y;
    Point(double a = 0, double b = 0){ x = a; y = b; }
};
//线段结构类型
struct LineSeg{
    Point s, e;
    LineSeg(){};
    LineSeg(Point a, Point b) : s(a), e(b){}
};
struct Line{
    double a, b, c;
};
Point operator-(Point a, Point b){
    return Point(a.x - b.x, a.y - b.y);
}
//重载==,判断点a,b是否相等
bool operator==(Point a, Point b){
    return abs(a.x - b.x) < EPS&&abs(a.y - b.y) < EPS;
}
//比较实数r1与r2的大小关系
int RlCmp(double r1, double r2 = 0){
    if (abs(r1 - r2) < EPS)
        return 0;
    return r1>r2 ? 1 : -1;
}
//返回向量p1-p0和p2-p0的叉积
double Cross(Point p0, Point p1, Point p2){
    Point a = p1 - p0;
    Point b = p2 - p0;
    return a.x*b.y - b.x*a.y;
}
//判断线段L1与线段L2是否相交(包括交点在线段上)
bool Intersect(LineSeg L1, LineSeg L2){
    //排斥实验和跨立实验
    return max(L1.s.x, L1.e.x) >= min(L2.s.x, L2.e.x)
        && min(L1.s.x, L1.e.x) <= max(L2.s.x, L2.e.x)
        && max(L1.s.y, L1.e.y) >= min(L2.s.y, L2.e.y)
        && min(L1.s.y, L1.e.y) <= max(L2.s.y, L2.e.y)
        && Cross(L1.s, L1.e, L2.s)*Cross(L1.s, L1.e, L2.e) <= 0
        && Cross(L2.s, L2.e, L1.s)*Cross(L2.s, L2.e, L1.e) <= 0;
}
Line MakeLine(Point a, Point b){
    Line L;
    L.a = (b.y - a.y);
    L.b = (a.x - b.x);
    L.c = (b.x*a.y - a.x*b.y);
    if (L.a < 0){  //保准x系数大于等于0
        L.a = -L.a;
        L.b = -L.b;
        L.c = -L.c;
    }
    return L;
}
//判直线X,Y是否相交,相交返回true和交点
bool LineIntersect(Line X, Line Y, Point&P){
    double d = X.a*Y.b - Y.a*X.b;
    if (d == 0) //直线平行或者重合
        return false;
    P.x = (X.b*Y.c - Y.b*X.c) / d;
    P.y = (X.c*Y.a - Y.c*X.a) / d;
    return true;
}
Point HighPoint(Point a, Point b){
    return  a.y > b.y ? a : b;
}
Point LowPoint(Point a, Point b){
    return a.y < b.y ? a : b;
}
double Dot(Point a, Point b){
    return a.x*b.x + a.y*b.y;
}
double Area(LineSeg L1, LineSeg L2){
    if (!Intersect(L1, L2))
        return 0;            //不相交
    Line X = MakeLine(L1.s, L1.e);
    Line Y = MakeLine(L2.s, L2.e);
    if (RlCmp(X.a*Y.a) == 0)  //有一条直线与x平行
        return 0;
    Point inter;
    LineIntersect(X, Y, inter);  //计算交点
    double y = min(max(L1.e.y, L1.s.y), max(L2.e.y, L2.s.y)); 
    double x1 = -(X.b*y + X.c) /X.a;
    double x2 = -(Y.b*y + Y.c) / Y.a;
    Point p1(x1, y), p2(x2, y);
    double area = abs(Cross(inter, p1, p2) / 2); //计算面积
    if (RlCmp(X.b*Y.b) == 0)  //有一条线与Y轴平行
        return area;      
    double k1 = -X.a / X.b;    //X的斜率
    double k2 = -Y.a / Y.b;    //Y的斜率
    Point high = HighPoint(HighPoint(L1.s, L1.e),HighPoint(L2.s, L2.e));
    Point low = LowPoint(HighPoint(L1.s, L1.e), HighPoint(L2.s, L2.e));
    if (RlCmp(high.y - low.y) == 0)
        return area;
    if (RlCmp(high.x - low.x) == 0)
        return 0;
    double k = -(high.x - low.x) / (high.y - low.y);
    Point mid((high.x + low.x) / 2, (high.y + low.y) / 2);
    Point op = mid - inter;
    Point ans;
    if (RlCmp(Dot(op, Point(1, k))) >= 0)
        ans = Point(1, k);
    else
        ans = Point(-1, -k);
    if ((ans.x > 0 && ans.y > 0 || (ans.x<0 && ans.y>0)))
        return area;
    else
        return 0;
}
int main(){
    int T;
    double a, b, c, d;
    scanf("%d", &T);
    while (T--){
        scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
        LineSeg L1(Point(a, b), Point(c, d));
        scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
        LineSeg L2(Point(a, b), Point(c, d));
        double area = Area(L1, L2);
        printf("%.2lf
", area);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/td15980891505/p/5735295.html