<h1 id="heading">题目</h1>
{% fold 点击显/隐题目 %}
<div class="oj"><div class="part" title="Description">
After counting so many stars in the sky in his childhood, Isaac, now an astronomer and a mathematician uses a big astronomical telescope and lets his image processing program count stars. The hardest part of the program is to judge if shining object in the sky is really a star. As a mathematician, the only way he knows is to apply a mathematical definition of stars.The mathematical definition of a star shape is as follows: A planar shape F is star-shaped if and only if there is a point C ∈ F such that, for any point P ∈ F, the line segment CP is contained in F. Such a point C is called a center of F. To get accustomed to the definition let’s see some examples below.
The first two are what you would normally call stars. According to the above definition, however, all shapes in the first row are star-shaped. The two in the second row are not. For each star shape, a center is indicated with a dot. Note that a star shape in general has infinitely many centers. Fore Example, for the third quadrangular shape, all points in it are centers.Your job is to write a program that tells whether a given polygonal shape is star-shaped or not.
</div><div class="part" title="Input">
The input is a sequence of datasets followed by a line containing a single zero. Each dataset specifies a polygon, and is formatted as follows.
n 
x1	y1 
x2	y2 
…
xn	yn
The first line is the number of vertices, n, which satisfies 4 ≤ n ≤ 50. Subsequent n lines are the x- and y-coordinates of the n vertices. They are integers and satisfy 0 ≤ xi ≤ 10000 and 0 ≤ yi ≤ 10000 (i = 1, …, n). Line segments (xi, yi)–(xi + 1, yi + 1) (i = 1, …, n − 1) and the line segment (xn, yn)–(x1, y1) form the border of the polygon in the counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions.You may assume that the polygon is simple, that is, its border never crosses or touches itself. You may assume assume that no three edges of the polygon meet at a single point even when they are infinitely extended.
</div><div class="part" title="Output">
For each dataset, output “1” if the polygon is star-shaped and “0” otherwise. Each number must be in a separate line and the line should not contain any other characters.
</div><div class="samp"><div class="clear"></div><div class="input part" title="Sample Input">
6 
66 13 
96 61 
76 98 
13 94 
4 0 
45 68 
8
27 21 
55 14 
93 12 
56 95 
15 48 
38 46 
51 65 
64 31 
0
</div><div class="output part" title="Sample Output">
1
0
</div><div class="clear"></div></div></div>
{% endfold %}

<h1 id="heading-1">题解</h1>
半平面交模板题,数据比较水
<h1 id="heading-2">代码</h1>
{% fold 点击显/隐代码 %}```cpp How I Mathematician Wonder What You Are! <a href="https://github.com/OhYee/sourcecode/tree/master/ACM">https://github.com/OhYee/sourcecode/tree/master/ACM</a> 代码备份 
#include <algorithm> 
#include <cmath> 
#include <cstdio> 
#include <cstring>
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 &amp;x) { 
if (fabs(x) &lt; eps) 
return 0; 
return x &gt; 0 ? 1 : -1; 
} 
struct Vector; 
inline double Cross(const Vector &amp;a, const Vector &amp;b);
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 &amp;rhs) const { 
return sgn(x - rhs.x) == 0 &amp;&amp; sgn(y - rhs.y) == 0; 
} 
bool operator!=(const Vector &amp;rhs) const { return !(this == rhs); } 
bool operator&lt;(const Vector &amp;rhs) const { 
if (sgn(y - rhs.y) == 0) 
return sgn(x - rhs.x) &lt; 0; 
return sgn(y - rhs.y) &gt; 0; 
} 
Vector operator+(const Vector &amp;rhs) const { 
return Vector(x + rhs.x, y + rhs.y); 
} 
Vector operator-(const Vector &amp;rhs) const { 
return Vector(x - rhs.x, y - rhs.y); 
} 
Vector operator(const double &amp;rhs) const { 
return Vector(x * rhs, y * rhs); 
} 
Vector operator/(const double &amp;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(&quot;(%.2f %.2f)&quot;, x, y); 
if (flag) 
Log(&quot;\n&quot;); 
} 
}; 
typedef Vector Point; 
struct Segment { 
Point a, b; 
Segment() {} 
Segment(Point _a, Point _b) : a(_a), b(_b) {} 
bool operator&lt;(const Segment &amp;rhs) const { 
double angle1 = getAngle(); 
double angle2 = rhs.getAngle(); 
if (sgn(angle1 - angle2) == 0) 
return sgn(Cross(rhs.toVector(), a)) &gt; 0; 
return sgn(angle1 - angle2) &lt; 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(&quot; -&gt; &quot;); 
b.print(); 
if (flag) 
Log(&quot;\n&quot;); 
} 
};
/** 
* 读入一个点的坐标 
* @return 读入的点 
*/ 
inline Point read_Point() { 
double x, y; 
scanf(&quot;%lf%lf&quot;, &amp;x, &amp;y); 
return Point(x, 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 &amp;a, const Vector &amp;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 &amp;a, const Vector &amp;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 &amp;a, const Point &amp;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 &amp;p, const Segment &amp;L) { 
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 &amp;L1, const Vector &amp;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 &amp;L1, const Segment &amp;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()) &gt; 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 &lt; 0 || t &gt; 1 || s &lt; 0 || s &gt; 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])) &gt;= 0; 
for (int i = 1; i &lt; numberOfSide &amp;&amp; ok; ++i) { 
if (!(Point_Segment(p, Segment(polygon[i - 1], polygon[i])) &gt;= 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) { 
double area = 0.0; 
for (int i = 2; i &lt; 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 &lt; numOfPoint; ++i) { 
if (!vis[p[i].n]) { 
while (pos &gt; 0 &amp;&amp; 
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) &gt; 0) { 
vis[ans[pos--].n] = false; 
} 
ans[++pos] = p[i]; 
vis[ans[pos].n] = true; 
} 
} 
// 左链 
for (int i = numOfPoint - 2; i &gt; 0; --i) { 
if (!vis[p[i].n]) { 
while (pos &gt; 0 &amp;&amp; 
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) &gt; 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(需要g++编译器) 
* @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 
*/ 
//用于计算的双端队列
//获取半平面交的多边形（多边形的核） 
//参数：向量集合[l]，向量数量[ln];(半平面方向在向量左边） 
//函数运行后如果n[即返回多边形的点数量]为0则不存在半平面交的多边形（不存在区域或区域面积无穷大） 
int Half_Plane(Segment s[], int numOfSide, Point ans[]) { 
sort(s, s + numOfSide);
<pre><code>int del = 0;
for (int i = 1; i &lt; numOfSide; ++i) {
 if (sgn(s[i].getAngle() - s[del].getAngle()) != 0)
 s[++del] = s[i];
}
numOfSide = 1 + del;

Segment dequeue[numOfSide * 2];
int bot = 0, top = 1;
dequeue[0] = s[0];
dequeue[1] = s[1];
for (int i = 2; i &lt; numOfSide; i++) {
 if (parallel(dequeue[top].toVector(), dequeue[top - 1].toVector()) ||
 parallel(dequeue[bot].toVector(), dequeue[bot + 1].toVector()))
 return 0;

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

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

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

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

return len;
</code></pre>
} 
// int Half_Plane3(Segment s[], int numOfSide, Point ans[]) { 
// sort(s, s + numOfSide);
// int del = 1; 
// for (int i = 1; i &lt; numOfSide; ++i) { 
// if (sgn(s[i].getAngle() - s[del].getAngle()) != 0) 
// s[del++] = s[i]; 
// } 
// numOfSide = del;
// int first, last; 
// Point *p = new Point[numOfSide]; 
// Segment *q = new Segment[numOfSide];
// q[first = last = 0] = s[0];
// for (int i = 1; i &lt; numOfSide; i++) { 
// while (first &lt; last &amp;&amp; Point_Segment(p[last - 1], s[i]) &lt; 0) 
// last--; 
// while (first &lt; last &amp;&amp; Point_Segment(p[first], s[i]) &lt; 0) 
// first++; 
// q[++last] = s[i];
// if (first &lt; last) 
// GetIntersection(q[last - 1], q[last], &amp;p[last - 1]); 
// } 
// while (first &lt; last &amp;&amp; Point_Segment(p[last - 1], q[first]) &lt; 0) 
// last--; 
// if (last - first &lt;= 1) 
// return 0;
// GetIntersection(q[last], q[first], &amp;p[last]);
// int m = 0; 
// for (int i = first; i &lt;= last; i++) 
// ans[m++] = p[i]; 
// return m; 
// } 
// int Half_Plane2(Segment s[], int numOfSide, Point ans[]) { 
// // s[numOfSide++] = Segment(Point(INF, INF), Point(-INF, INF)); 
// // s[numOfSide++] = Segment(Point(-INF, INF), Point(-INF, -INF)); 
// // s[numOfSide++] = Segment(Point(-INF, -INF), Point(INF, -INF)); 
// // s[numOfSide++] = Segment(Point(INF, -INF), Point(INF, INF)); 
// sort(s, s + numOfSide);
// Log(&quot;\n\nHalf_Plane\n&quot;);
// int del = 0; 
// for (int i = 1; i &lt; numOfSide; ++i) { 
// if (sgn(s[i].getAngle() - s[del].getAngle()) != 0) 
// s[++del] = s[i]; 
// } 
// numOfSide = del + 1;
// for (int i = 0; i &lt; numOfSide; ++i) 
// s[i].print(1);
// int front = -1, rear = 0; 
// Segment q[numOfSide];
// for (int i = 0; i &lt; numOfSide; ++i) { 
// int it = i % numOfSide;
// Log(&quot;%d Testing&quot;, i); 
// s[it].print(1);
// while (front &gt; rear &amp;&amp; Point_Segment(ans[front], s[it]) &lt; 0) { 
// front--; 
// Log(&quot;\tHead pop\n&quot;); 
// }
// while (front &gt; rear &amp;&amp; Point_Segment(ans[rear + 1], s[it]) &lt; 0) { 
// rear++; 
// Log(&quot;\tRear pop\n&quot;); 
// }
// if (front - rear &lt; 0) { 
// q[++front] = s[it]; 
// Log(&quot;Add [%d]&quot;, front); 
// s[it].print(1); 
// } else { 
// Point p; 
// int po = Segment_Segment(s[it], q[front], &amp;p); 
// // 
// Log(&quot;getpoint(&quot;),s[it].print(),q[front].print(),Log(&quot;)=&quot;),p.print(1); 
// if (po == 0 || po == 1) { 
// ans[++front] = p; 
// q[front] = s[it]; 
// Log(&quot;Add [%d]&quot;, front); 
// p.print(0); 
// Log(&quot; [%d]&quot;, front); 
// s[it].print(1); 
// } else { 
// Log(&quot;ERROR\n&quot;); 
// } 
// }
// Log(&quot;front:%d rear:%d\t&quot;, front, rear); 
// for (int j = rear + 1; j &lt;= front; ++j) { 
// ans[j].print(); 
// Log(&quot; &quot;); 
// } 
// Log(&quot;\n\n&quot;); 
// }
// Log(&quot;front:%d rear:%d\n&quot;, front, rear); 
// int len = front - rear; 
// for (int i = 0; i &lt; len; ++i) 
// ans[i] = ans[rear + i + 1]; 
// for (int i = 1; i &lt; len; ++i) 
// if (ans[i] == ans[i - 1]) { 
// ans[i] = ans[i + 1]; 
// len--; 
// } 
// return len; 
// }
const int maxn = 1505; 
bool vis[maxn]; 
Point p[maxn]; 
Point ans[maxn]; 
Segment s[maxn];
int main() { 
int n; 
while (scanf(&quot;%d&quot;, &amp;n), n != 0) { 
for (int i = 0; i &lt; n; ++i) { 
p[i] = read_Point(); 
} 
for (int i = 0; i &lt; n; ++i) 
s[i] = Segment(p[i], p[(i + 1) % n]);
<pre><code> int len = Half_Plane(s, n, ans);

 printf(&quot;%.7f\n&quot;, getArea(ans, len));
}
return 0;
</code></pre>
}
<pre style="background-color:#fff">{% endfold %}</pre>

POJ 3130.How I Mathematician Wonder What You Are!