题目

{% fold 点击显/隐题目 %}

A rectangular area specified in world coordinates is called a window. The rectangular area on the display device to which a window is mapped is called a viewport. The picture elements that fall within this window area is visible. In the viewport there is a block which is given higher priority for displaying. That is, those line segments that falls behind the block will not be visible. Assume the boundaries of a viewport: $x=x\_{min}$, $x=x\_{max}$, $y=y\_{min}$, $y=y\_{max}$, the boundaries of block: $x=x\_{min}$, $x=x\_{max}$, $y=y\_{min}$, $y=y\_{max}$, and $(x\_{i1}\ , y\_{i1} )$ $(x\_{i2}\ , y\_{i2})$ : he endpoints of the lines in device coordinates. ![](/post/img/jisuanke17318.png) Please calculate the coordinate value of the endpoints of all the visible line.
The boundaries of a viewport and block, endpoints of the lines in device coordinates. $x\_{min}$ $y\_{min}$ $x\_{max}$ $y\_{max}$ $x\_{min}$ $y\_{min}$ $x\_{max}$ $y\_{max}$ $x\_{11}$ $y\_{11}$ $x\_{12}$ $y\_{12}$ $x\_{21}$ $y\_{21}$ $x\_{22}$ $y\_{22}$ $...$ $x\_{n1}$ $y\_{n1}$ $x\_{n2}$ $y\_{n2}$
Endpoints of all the visible lines. Please sort the coordinate values of $x\_{i1}$ ’s in descending order. $x\_{11}$ $y\_{11}$ $x\_{12}$ $y\_{12}$ $x\_{21}$ $y\_{21}$ $x\_{22}$ $y\_{22}$ $...$ $x\_{k1}$ $y\_{k1}$ $x\_{k2}$ $y\_{k2}$ Note: 1. $x\_{i1}\ <\ x\_{i2}$ 2. In the condition of only the endpoints of the segments are falling on the boundaries of the viewport or block and the other part is invisible, please neglect it. 3. Integers: $x\_{min}$, $y\_{min}$ , $x\_{max}$ , $y\_{max}$, $x\_{min}$ , $y\_{min}$ , $x\_{max}$ , $y\_{max}$ ; float: $x\_{ij}$, $y\_{ij}$. 4. All float number should be accurate to $10^{-4}$
1 2 2 3 3.0 3.0 7.0 7.0 9.0 7.0 2.0 6.0 4.0 2.0 1.1 4.0 9.0 -1.0 5.1 7.0 2.0 2.0 1.0 3.0 -2.5 3.1 5.2 -3.0 -1.0 7.3 -2.1 0.6 -2.1 5.6 6.0 7.0 -0.7 2.1 -5.2 12.6
3.0000 3.0000 5.0000 5.0000 1.1000 4.0000 4.0000 2.0000 0.0000 1.1195 1.4131 0.0000
{% endfold %}

题解

给你不超过50个线段,有一个大矩形为可视区域,一个小矩形为遮挡区域,输出所有能够看到的线段的首尾坐标(线段可以被小矩形分成两部分或者只能看到原线段的一部分)

按照讨论区的说法,如果正好在边界上,认为是不可见
根据线段的两个端点与矩形的关系以及与矩形的交点,可以分为如下情况:

  1. 整条线段都在矩形外部(保留整个线段 | 删除整个线段)
  2. 整条线段都在矩形上(删除整个线段)
  3. 整条线断都在矩形内部(删除整个线段 | 保留整个线段)
  4. 两个端点在矩形外部,但是和矩形有一个交点(分为两个线段 | 删除整个线段)
  5. 两个端点在矩形外部,但是和矩形有两个交点(分为两个线段 | 变成一个新线段)
  6. 一个端点在矩形内部,另一个在外部或者矩形上(变成一个新线段)

然后分别对应两种矩形进行不同的操作

最后按照要求排序输出即可

{% note danger%}
注意最后输出可能会有 -0.0000
输出时使用 fabs()
{% endnote %}

代码

{% fold 点击显/隐代码 %}```cpp Line Segments Clipped by Windows https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
#include
#include
using namespace std;
#define Log(format, ...) // printf(format, ##VA_ARGS)
/* 计算几何模板 */
const double eps = 1e-10;
const double INF = 0x4242424242424242;
inline int sgn(const double &x) {
if (fabs(x) < eps)
return 0;
return x > 0 ? 1 : -1;
}
struct Vector;
struct Segment;
double Cross(const Vector &, const Vector &);
int Point_Segment(const Vector &, const Segment &);
struct Vector {
double x, y;
int n;
Vector(double _x = 0, double _y = 0, int _n = 0) : x(_x), y(_y), n(_n) {}
bool operator==(const Vector &rhs) const {
return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0;
}
bool operator!=(const Vector &rhs) const { return !(*this == rhs); }
bool operator<(const Vector &rhs) const {
if (sgn(x - rhs.x) == 0)
return sgn(y - rhs.y) < 0;
return sgn(x - rhs.x) < 0;
}
Vector operator+(const Vector &rhs) const {
return Vector(x + rhs.x, y + rhs.y);
}
Vector operator-(const Vector &rhs) const {
return Vector(x - rhs.x, y - rhs.y);
}
Vector operator*(const double &rhs) const {
return Vector(x * rhs, y * rhs);
}
Vector operator/(const double &rhs) const {
return Vector(x / rhs, y / rhs);
}
double getAngle() { return atan2(y, x); }
double squre() const { return x * x + y * y; }
double distance() const { return sqrt(squre()); }
void print(bool flag = 0) const {
Log("(%.2f %.2f)", x, y);
if (flag)
Log("\n");
}
};
typedef Vector Point;
struct Segment {
Point a, b;
Segment() {}
Segment(Point _a, Point _b) : a(_a), b(_b) {}
bool operator<(const Segment &rhs) const {
double angle1 = getAngle();
double angle2 = rhs.getAngle();
if (sgn(angle1 - angle2) == 0)
return Point_Segment(a, rhs) > 0;
return sgn(angle1 - angle2) < 0;
}
double getAngle() const { return toVector().getAngle(); }
Vector toVector() const { return b - a; }
double distance() const { return toVector().distance(); }
void print(bool flag = 0) const {
a.print();
Log(" -> ");
b.print();
if (flag)
Log("\n");
}
};
/**
* 读入一个点的坐标
* @return 读入的点
*/
inline Point read_Point() {
double x, y;
scanf("%lf%lf", &x, &y);
return Point(x, y);
}
inline double xmult(const Vector &a, const Vector &b, const Vector &c) {
return (a.x - c.x) * (b.y - c.y) - (b.x - c.x) * (a.y - c.y);
}
/**
* 计算两个向量的叉积
* @param a 向量1
* @param b 向量2
* @return 叉积
*/
inline double Cross(const Vector &a, const Vector &b) {
return a.x * b.y - a.y * b.x;
}
/**
* 计算两个向量的点积
* @param a 向量1
* @param b 向量2
* @return 点积
*/
inline double Dot(const Vector &a, const Vector &b) {
return a.x * b.x + a.y * b.y;
}
/**
* 计算两点之间的距离
* @param a 线段L1
* @param b 线段L2
* @return 两点间的距离
*/
inline double Distance(const Point &a, const Point &b) {
return (a - b).distance();
}
/**
* 点和直线的关系
* @param p 目标点
* @param L 目标直线
* @return 1 在左侧,0 在直线上,-1在右侧
*/
inline int Point_Segment(const Vector &p, const Segment &L) {
// printf("Point_segment %d\n", sgn(Cross(L.b - L.a, p - L.a)));
return sgn(Cross(L.b - L.a, p - L.a));
}
/**
* 计算两个线段(直线)是否平行
* @param L1 L1的向量
* @param L2 L2的向量
* @return 返回是否平行
*/
bool parallel(const Vector &L1, const Vector &L2) {
return sgn(Cross(L1, L2)) == 0;
}
/**
* 计算两个直线的交点(需要确保不平行、重合)
* @param L1 L1的向量
* @param L2 L2的向量
* @return 返回是否平行
*/
Point getIntersection(const Segment &L1, const Segment &L2) {
Point ret = L1.a;
double t = ((L1.a.x - L2.a.x) * (L2.a.y - L2.b.y) -
(L1.a.y - L2.a.y) * (L2.a.x - L2.b.x)) /
((L1.a.x - L1.b.x) * (L2.a.y - L2.b.y) -
(L1.a.y - L1.b.y) * (L2.a.x - L2.b.x));
ret.x += (L1.b.x - L1.a.x) * t;
ret.y += (L1.b.y - L1.a.y) * t;
return ret;
}
/**
* 计算两个线段的位置关系
* @param L1 线段L1
* @param L2 线段L2
* @param p 返回交点坐标
* @return 2 重叠
1 相交
0 延长线相交
-1 平行
-2 共线不交
*/
inline int Segment_Segment(const Segment &L1, const Segment &L2,
Point *p = NULL) {
double a = L1.b.x - L1.a.x;
double b = L2.b.x - L2.a.x;
double c = L1.b.y - L1.a.y;
double d = L2.b.y - L2.a.y;
double f = a * d - b * c;
// 平行或重叠
if (sgn(f) == 0) {
if (Point_Segment(L1.a, L2)) {
// 平行
return -1;
} else {
// 共线
int len = max(max(Distance(L1.a, L2.a), Distance(L1.a, L2.b)),
max(Distance(L1.b, L2.a), Distance(L1.b, L2.b)));
if (sgn(len - L1.distance() - L2.distance()) > 0) {
// 共线不交
return -2;
} else {
// 重叠
return 2;
}
}
}
double g = L2.b.x - L1.a.x;
double h = L2.b.y - L1.a.y;
double t = (d * g - b * h) / f;
double s = (-c * g + a * h) / f;
if (p != NULL)
*p = Point(L1.a.x + t * a, L1.a.y + t * c);
// 在延长线上
if (t < 0 || t > 1 || s < 0 || s > 1)
return 0;
// 线段相交
return 1;
}
/**
* 判断点是否在多边形内部
* @param p 需要判断的点
* @param polygon 多边形点集,需要保证有序
* @param numberOfSide 多边形边数
* @return 1 点在多边形内,0在多边形上, -1 点不在多边形内
*/
int Point_Polygon(const Point &p, const Point polygon[],
const int numberOfSide) {
int res = Point_Segment(p, Segment(polygon[numberOfSide - 1], polygon[0]));
for (int i = 1; i < numberOfSide; ++i)
res = min(res, Point_Segment(p, Segment(polygon[i - 1], polygon[i])));
return res;
}

const int maxn = 505;
Point Rec1[4], Rec2[4];
Segment s[maxn];
bool del[maxn];
int n;

// 保留里面的
bool judge1(int t, Point Rec[]) {
bool era = false;

int in[2]; // 两个点是否在矩形内
in[0] = Point_Polygon(s[t].a, Rec, 4);
in[1] = Point_Polygon(s[t].b, Rec, 4);

if (in[0] == 0 && in[1] == 0) {
    // 正好在矩形上
    era = true;
} else if (in[0] >= 0 && in[1] >= 0) {
    // 在矩形内部
    era = false;
} else {
    // 交点个数
    int cnt = 0;
    Point p, pp[2];
    for (int i = 0; i < 4; i++) {
        if (1 ==
            Segment_Segment(s[t], Segment(Rec[i], Rec[(i + 1) % 4]), &p)) {
            pp[cnt++] = p;
        }
    }
    if (cnt == 2 && pp[0] == pp[1])
        --cnt;

    if (in[0] < 0 && in[1] < 0) {
        // 两个点都在矩形外部
        if (cnt == 0) {
            // 全部在矩形外部
            era = true;
        } else {
            if (cnt == 1) {
                // 有一个交点
                era = true;
            } else {
                // 有两个交点
                s[n++] = Segment(pp[0], pp[1]);
                era = true;
            }
        }
    } else {
        // 一半在外面,一半在里面
        era = true;
        if (in[0] >= 0) {
            s[n++] = Segment(pp[0], s[t].a);
        } else {
            s[n++] = Segment(pp[0], s[t].b);
        }
    }
}
del[t] = era;
return era;

}

// 保留外面的
bool judge2(int t, Point Rec[]) {
bool era = false;

int in[2]; // 两个点是否在矩形内
in[0] = Point_Polygon(s[t].a, Rec, 4);
in[1] = Point_Polygon(s[t].b, Rec, 4);

if (in[0] == 0 && in[1] == 0) {
    // 正好在矩形上
    era = true;
} else if (in[0] >= 0 && in[1] >= 0) {
    // 在矩形内部
    era = true;
} else {
    // 交点个数
    int cnt = 0;
    Point p, pp[2];
    for (int i = 0; i < 4; i++) {
        if (1 ==
            Segment_Segment(s[t], Segment(Rec[i], Rec[(i + 1) % 4]), &p)) {
            pp[cnt++] = p;
        }
    }
    if (cnt == 2 && pp[0] == pp[1])
        --cnt;

    if (in[0] < 0 && in[1] < 0) {
        // 两个点都在矩形外部
        if (cnt == 0) {
            // 全部在矩形外部
            era = false;
        } else {
            if (cnt == 1) {
                // 有一个交点
                s[n++] = Segment(pp[0], s[t].a);
                s[n++] = Segment(pp[0], s[t].b);
                era = true;
            } else {
                // 有两个交点
                if (Distance(pp[0], s[t].a) + eps <
                    Distance(pp[1], s[t].b)) {
                    s[n++] = Segment(pp[0], s[t].a);
                    s[n++] = Segment(pp[1], s[t].b);
                } else {
                    s[n++] = Segment(pp[1], s[t].a);
                    s[n++] = Segment(pp[0], s[t].b);
                }

                era = true;
            }
        }
    } else {
        // 一半在外面,一半在里面
        era = true;
        if (in[0] >= 0) {
            s[n++] = Segment(pp[0], s[t].b);
        } else {
            s[n++] = Segment(pp[0], s[t].a);
        }
    }
}
del[t] = era;
return era;

}

void Erase() {
int pos = 0, cnt = 0;
for (int i = 0; i < n; ++i) {
if (del[i]) {
++cnt;
continue;
} else {
s[pos++] = s[i];
}
}
n -= cnt;
memset(del, false, sizeof(del));
}

void cmp1(Segment &t) {
if (t.b < t.a)
swap(t.a, t.b);
}
bool cmp2(Segment a, Segment b) {
if (a.a == b.a)
return b.b < a.b;
return b.a < a.a;
}

int main() {
Rec1[0] = read_Point();
Rec1[2] = read_Point();
Rec1[1] = Point(Rec1[2].x, Rec1[0].y);
Rec1[3] = Point(Rec1[0].x, Rec1[2].y);

Rec2[0] = read_Point();
Rec2[2] = read_Point();
Rec2[1] = Point(Rec2[2].x, Rec2[0].y);
Rec2[3] = Point(Rec2[0].x, Rec2[2].y);

n = 0;
memset(del, false, sizeof(del));

double x1, x2, y1, y2;
while (~scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2))
    s[n++] = Segment(Point(x1, y1), Point(x2, y2));

// for (int i = 0; i < n; ++i)
//     printf("%.4f %.4f %.4f %.4f\n", s[i].a.x, s[i].a.y, s[i].b.x,
//     s[i].b.y);
// printf("\n");

int oldn = n;
for (int i = 0; i < oldn; ++i)
    judge1(i, Rec1);
Erase();

// for (int i = 0; i < n; ++i)
//     printf("%.4f %.4f %.4f %.4f\n", s[i].a.x, s[i].a.y, s[i].b.x,
//     s[i].b.y);
// printf("\n");

oldn = n;
for (int i = 0; i < oldn; ++i)
    judge2(i, Rec2);
Erase();

// for (int i = 0; i < n; ++i)
//     printf("%.4f %.4f %.4f %.4f\n", s[i].a.x, s[i].a.y, s[i].b.x,
//     s[i].b.y);
// printf("\n");

for (int i = 0; i < n; ++i)
    cmp1(s[i]);
sort(s, s + n, cmp2);

for (int i = 0; i < n; ++i)
    printf("%.4f %.4f %.4f %.4f\n", fabs(s[i].a.x), fabs(s[i].a.y),
           fabs(s[i].b.x), fabs(s[i].b.y));

return 0;

}

{% endfold %}