给出一个多边形，要求判断它的内核是否存在。
还是半平面交的题，在这道题中，公告板允许其所在位置与直线共线也算是可见，于是我们就可以将每一条直线微小的移动，然后判断是够能够交出多边形，这样做是因为对于半平面交是不能直接判断是够交集是一个点的情况的。
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath> using namespace std; struct Point {
double x, y;
Point() {}
Point(double x, double y) : x(x), y(y) {}
} ;
template<class T> T sqr(T x) { return x * x;}
typedef Point Vec;
Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);}
Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);}
Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);}
Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);} const double EPS = 1e-;
const double PI = acos(-1.0);
inline int sgn(double x) { return (x > EPS) - (x < -EPS);} inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}
inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}
inline double dotDet(Point o, Point a, Point b) { return dotDet(a - o, b - o);}
inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}
inline double vecLen(Vec x) { return sqrt(dotDet(x, x));}
inline double toRad(double deg) { return deg / 180.0 * PI;}
inline double angle(Vec v) { return atan2(v.y, v.x);}
inline Vec vecUnit(Vec x) { return x / vecLen(x);}
inline Vec normal(Vec x) { return Vec(-x.y, x.x) / vecLen(x);} const int N = ;
struct DLine {
Point p;
Vec v;
double ang;
DLine() {}
DLine(Point p, Vec v) : p(p), v(v) { ang = atan2(v.y, v.x);}
bool operator < (DLine L) const { return ang < L.ang;}
DLine move(double x) {
Vec nor = normal(v);
nor = nor * x;
return DLine(p + nor, v);
}
} dl[N];
Point pt[N]; inline bool onLeft(DLine L, Point p) { return crossDet(L.v, p - L.p) > ;}
Point dLineIntersect(DLine a, DLine b) {
Vec u = a.p - b.p;
double t = crossDet(b.v, u) / crossDet(a.v, b.v);
return a.p + a.v * t;
} struct Poly {
vector<Point> pt;
Poly() { pt.clear();}
~Poly() {}
Poly(vector<Point> &pt) : pt(pt) {}
Point operator [] (int x) { return pt[x];}
int size() { return pt.size();}
double area() {
double ret = 0.0;
int sz = pt.size();
pt.push_back(pt[]);
for (int i = ; i <= sz; i++) ret += crossDet(pt[i], pt[i - ]);
pt.pop_back();
return fabs(ret / 2.0);
}
} ; Poly halfPlane(DLine *L, int n) {
Poly ret = Poly();
sort(L, L + n);
int fi, la;
Point *p = new Point[n];
DLine *q = new DLine[n];
q[fi = la = ] = L[];
for (int i = ; i < n; i++) {
while (fi < la && !onLeft(L[i], p[la - ])) la--;
while (fi < la && !onLeft(L[i], p[fi])) fi++;
q[++la] = L[i];
if (sgn(crossDet(q[la].v, q[la - ].v)) == ) {
la--;
if (onLeft(q[la], L[i].p)) q[la] = L[i];
}
if (fi < la) p[la - ] = dLineIntersect(q[la - ], q[la]);
}
while (fi < la && !onLeft(q[fi], p[la - ])) la--;
if (la <= fi) return ret;
p[la] = dLineIntersect(q[la], q[fi]);
for (int i = fi; i <= la; i++) ret.pt.push_back(p[i]);
return ret;
} bool isClockwise(Point *pt, int n) {
double sum = 0.0;
pt[n] = pt[];
Point O = Point(0.0, 0.0);
for (int i = ; i < n; i++) {
sum += crossDet(O, pt[i], pt[i + ]);
}
return sum < ;
} int main() {
// freopen("in", "r", stdin);
int T, n;
cin >> T;
while (T-- && cin >> n) {
for (int i = ; i < n; i++) cin >> pt[i].x >> pt[i].y;
pt[n] = pt[];
if (isClockwise(pt, n)) for (int i = ; i < n; i++) dl[i] = DLine(pt[i + ], pt[i] - pt[i + ]).move(-EPS);
else for (int i = ; i < n; i++) dl[i] = DLine(pt[i], pt[i + ] - pt[i]).move(-EPS);
Poly tmp = halfPlane(dl, n);
if (tmp.size() > ) puts("YES");
else puts("NO");
}
return ;
} /*
3
6
0 0
100 0
100 100
0 100
50 75
50 25
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
*/
——written by Lyon