POJ 2826 An Easy Problem?!（计算几何)

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基本就和网上题解一样的思路，把几种情况判掉blablabla。。就WA了。

然后答案加个EPS就过了。显然这题没有写SPJ。。卡了精度

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) {
        this->x = x;
        this->y = y;
    }
    void read() {
        scanf("%lf%lf", &x, &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);
}

bool operator < (const Point& a, const Point& b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

const double eps = 1e-8;

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积

struct Line {
    Point v, p;
    Line() {}
    Line(Point v, Point p) {
        this->v = v;
        this->p = p;
    }
    Point point(double t) {
        return v + p * t;
    }
};

//向量旋转
Vector Rotate(Vector A, double rad) {
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

Vector AngleBisector(Point p, Vector v1, Vector v2){//给定两个向量，求角平分线
    double rad = Angle(v1, v2);
    return Rotate(v1, dcmp(Cross(v1, v2)) * 0.5 * rad);
}

//判断3点共线
bool LineCoincide(Point p1, Point p2, Point p3) {
    return dcmp(Cross(p2 - p1, p3 - p1)) == 0;
}

//判断向量平行
bool LineParallel(Vector v, Vector w) {
    return Cross(v, w) == 0;
}

//判断向量垂直
bool LineVertical(Vector v, Vector w) {
    return Dot(v, w) == 0;
}

//计算两直线交点，平行，重合要先判断
Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
    Vector u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}

//点到直线距离
double DistanceToLine(Point P, Point A, Point B) {
    Vector v1 = B - A, v2 = P - A;
    return fabs(Cross(v1, v2)) / Length(v1);
}

//点到线段距离
double DistanceToSegment(Point P, Point A, Point B) {
    if (A == B) return Length(P - A);
    Vector v1 = B - A, v2 = P - A, v3 = P - B;
    if (dcmp(Dot(v1, v2)) < 0) return Length(v2);
    else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);
    else return fabs(Cross(v1, v2)) / Length(v1);
}

//点在直线上的投影点
Point GetLineProjection(Point P, Point A, Point B) {
    Vector v = B - A;
    return A + v * (Dot(v, P - A) / Dot(v, v));
}

//线段相交判定(规范相交)
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
    double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

//可以不规范相交
bool SegmentProperIntersection2(Point a1, Point a2, Point b1, Point b2) {
    double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
    return max(a1.x, a2.x) >= min(b1.x, b2.x) &&
    max(b1.x, b2.x) >= min(a1.x, a2.x) &&
    max(a1.y, a2.y) >= min(b1.y, b2.y) &&
    max(b1.y, b2.y) >= min(a1.y, a2.y) &&
    dcmp(c1) * dcmp(c2) <= 0 && dcmp(c3) * dcmp(c4) <= 0;
}

//判断点在线段上, 不包含端点
bool OnSegment(Point p, Point a1, Point a2) {
    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}

//n边形的面积
double PolygonArea(Point *p, int n) {
    double area = 0;
    for (int i = 1; i < n - 1; i++)
        area += Cross(p[i] - p[0], p[i + 1] - p[0]);
    return area / 2;
}

int t;
Point p[4];

double solve() {
    if (LineParallel(p[1] - p[0], p[3] - p[2])) return 0.0;
    if (!SegmentProperIntersection2(p[0], p[1], p[2], p[3])) return 0.0;
    Point x = GetLineIntersection(p[0], p[1] - p[0], p[2], p[3] - p[2]);
    if (p[0].y > x.y && p[2].y > x.y) {
        if ((p[0].x - x.x) * (p[2].x - x.x) > 0) {
            if (dcmp(fabs(p[0].x - x.x) - fabs(p[2].x - x.x)) >= 0) {
                Point sb = GetLineIntersection(x, p[0] - x, p[2], Vector(0, 1));
                if (sb.y > p[2].y)
                    return 0.0;
            }
        }
        Point tmp = GetLineIntersection(p[0], p[1] - p[0], p[2], Vector(1, 0));
        return fabs(Area2(p[2], x, tmp) / 2);
    }
    return 0.0;
}

int main() {
    scanf("%d", &t);
    while (t--) {
        for (int i = 0; i < 4; i++) p[i].read();
        if (p[0].y < p[1].y) swap(p[0], p[1]);
        if (p[2].y < p[3].y) swap(p[2], p[3]);
        if (p[0].y < p[2].y) {
            swap(p[0], p[2]);
            swap(p[1], p[3]);
        }
        printf("%.2f\n", solve() + eps);
    }
    return 0;
}

POJ 2826 An Easy Problem?!（计算几何)

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原文地址：http://blog.csdn.net/accelerator_/article/details/44979065