题目

{% fold 点击显/隐题目 %}

A friend of yours has taken the job of security officer at the Star-Buy Company, a famous depart- ment store. One of his tasks is to install a video surveillance system to guarantee the security of the customers (and the security of the merchandise of course) on all of the store's countless floors. As the company has only a limited budget, there will be only one camera on every floor. But these cameras may turn around to look in every direction.

The first problem is to choose where to install the camera for every floor. The only requirement is that every part of the room must be visible from there. In the following figure the left floor can be completely surveyed from the position indicated by a dot, while for the right floor, there is no such position, the given position failing to see the lower left part of the floor.

Before trying to install the cameras, your friend first wants to know whether there is indeed a suitable position for them. He therefore asks you to write a program that, given a ground plan, de- termines whether there is a position from which the whole floor is visible. All floor ground plans form rectangular polygons, whose edges do not intersect each other and touch each other only at the corners.

The input contains several floor descriptions. Every description starts with the number n of vertices that bound the floor (4 <= n <= 100). The next n lines contain two integers each, the x and y coordinates for the n vertices, given in clockwise order. All vertices will be distinct and at corners of the polygon. Thus the edges alternate between horizontal and vertical.

A zero value for n indicates the end of the input.

For every test case first output a line with the number of the floor, as shown in the sample output. Then print a line stating "Surveillance is possible." if there exists a position from which the entire floor can be observed, or print "Surveillance is impossible." if there is no such position.

Print a blank line after each test case.

4 0 0 0 1 1 1 1 0 8 0 0 0 2 1 2 1 1 2 1 2 2 3 2 3 0 0
Floor #1 Surveillance is possible.

Floor #2
Surveillance is impossible.

{% endfold %}

题解

半平面交模板题,和POJ 1279一样,数据比较严

代码

{% fold 点击显/隐代码 %}```cpp Video Surveillance https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
#include

using namespace std;

#define Log(format, ...) // printf(format, ##VA_ARGS)

/* 计算几何模板 */

const double eps = 1e-8;
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;
inline double Cross(const Vector &a, const Vector &b);
int Point_Segment(const Vector &p, const Segment &L);
inline double xmult(const Vector &a, const Vector &b, const Vector &c);

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(y - rhs.y) == 0)
return sgn(x - rhs.x) < 0;
return sgn(y - rhs.y) > 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) {
//注意先判断平行和重合
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 true 点在多边形内,false 点不在多边形内
*/
bool Point_Polygon(Point p, Point polygon[], int numberOfSide) {
bool ok =
Point_Segment(p, Segment(polygon[numberOfSide - 1], polygon[0])) >= 0;
for (int i = 1; i < numberOfSide && ok; ++i) {
if (!(Point_Segment(p, Segment(polygon[i - 1], polygon[i])) >= 0))
ok = false;
}
return ok;
}

/**
* 求多边形面积
* @param p 点集
* @param numOfSide 多边形边数
* @return 返回多边形面积
*/
double getArea(Point p[], int numberOfSide) {
if (numberOfSide < 3)
return 0.0;
double area = 0.0;
for (int i = 2; i < numberOfSide; ++i)
area += fabs(0.5 * Cross(p[i] - p[0], p[i - 1] - p[0]));
return area;
}

/**
* 求点集的凸包
* @param p 点集,需要保证已经排序,并且点需要有不同的编号
* @param numOfPoint 点集内的点的个数
* @param vis 记录点对应的编号是否可以选择
* @param ans 返回的凸包
* @param begin 起始位置
* @return 返回凸包上点的个数
*/
int Convex_Hull(Point p[], int numOfPoint, bool vis[], Point ans[],
int begin = 0) {
int pos = begin - 1;
// 右链
for (int i = 0; i < numOfPoint; ++i) {
if (!vis[p[i].n]) {
while (pos > 0 &&
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) > 0) {
vis[ans[pos--].n] = false;
}
ans[++pos] = p[i];
vis[ans[pos].n] = true;
}
}
// 左链
for (int i = numOfPoint - 2; i > 0; --i) {
if (!vis[p[i].n]) {
while (pos > 0 &&
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) > 0) {
vis[ans[pos--].n] = false;
}
ans[++pos] = p[i];
vis[ans[pos].n] = true;
}
}
return pos + 1;
}

/**
* 求直线集合的半平面交
* @param s 线段集合
* @param numOfSide 线段数目
* @param ans 返回的多边形点
* @return 返回结果多边形点的个数
*/
Segment dequeue[3000];
int Half_Plane(Segment s[], int numOfSide, Point ans[]) {
sort(s, s + numOfSide);

int del = 1;
for (int i = 1; i < numOfSide; ++i) {
    if (sgn(s[i].getAngle() - s[del - 1].getAngle()) != 0)
        s[del++] = s[i];
}
numOfSide = del;

for (int i = 0; i < numOfSide; ++i) {
    s[i].print(1);
}

int bot = 0, top = 1;
dequeue[0] = s[0];
dequeue[1] = s[1];
for (int i = 2; i < numOfSide; i++) {
    if (parallel(dequeue[top].toVector(), dequeue[top - 1].toVector()) ||
        parallel(dequeue[bot].toVector(), dequeue[bot + 1].toVector())) {
        dequeue[top].print(), dequeue[top - 1].print(1);
        dequeue[bot].print(), dequeue[bot + 1].print(1);
        return 0;
    }

    while (bot < top &&
           Point_Segment(getIntersection(dequeue[top], dequeue[top - 1]),
                         s[i]) < 0)
        top--;
    while (bot < top &&
           Point_Segment(getIntersection(dequeue[bot], dequeue[bot + 1]),
                         s[i]) < 0)
        bot++;
    dequeue[++top] = s[i];
}

while (bot < top &&
       Point_Segment(getIntersection(dequeue[top], dequeue[top - 1]),
                     dequeue[bot]) < 0)
    top--;
while (bot < top &&
       Point_Segment(getIntersection(dequeue[bot], dequeue[bot + 1]),
                     dequeue[top]) < 0)
    bot++;

//计算交点(注意不同直线形成的交点可能重合)
if (top <= bot + 1)
    return 0;

int len = 0;
for (int i = bot; i < top; i++)
    ans[len++] = getIntersection(dequeue[i], dequeue[i + 1]);
if (bot < top + 1)
    ans[len++] = getIntersection(dequeue[bot], dequeue[top]);

return len;

}

const int maxn = 1005;
bool vis[maxn];
Point p[maxn];
Point ans[maxn];
Segment s[maxn];

int main() {
int n;
int kase = 1;
while (scanf("%d", &n), n != 0) {
for (int i = 0; i < n; ++i) {
p[i] = read_Point();
p[i].n = i;
}
for (int i = 0; i < n; ++i)
s[i] = Segment(p[(i + 1) % n], p[i]);

    int len = Half_Plane(s, n, ans);

    // printf("%d\n",len);
    if (kase > 1)
        printf("\n");
    printf("Floor #%d\nSurveillance is %spossible.\n", kase++,
           len > 0 ? "" : "im");
}
return 0;

}

{% endfold %}