題目鏈接：點擊查看

題目大意：給出n個點，以及一個圓心和半徑，首先判斷這n個點能否構成凸包，若能，繼續判斷圓是否在凸包內

題目分析：這個題目確實非常綜合，考察了判斷凸包問題，判斷圓是否在凸包內，只要保證圓心在凸包內，以及圓心到每條邊的距離小于等于半徑即可，說到底還是考察了模板是否靠譜，不得不佩服kuangbin大神的模板真的牛，一套用下來沒有一點bug，直接貼代碼吧，全是模板，懶得刪減，所以有點長

代碼：

#include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<queue> #include<map> #include<set> #include<sstream> using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const int N=110;const double eps = 1e-8;const double pi = acos(-1.0);const int maxp = 110;int sgn(double x){if(fabs(x) < eps)return 0;if(x < 0)return -1;else return 1; }struct Point{double x,y;Point(){}Point(double _x,double _y){x = _x;y = _y;}void input(){scanf("%lf%lf",&x,&y);}void output(){printf("%.2f %.2f\n",x,y);}bool operator == (Point b)const{return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;}bool operator < (Point b)const{return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;}Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}//叉積double operator ^(const Point &b)const{return x*b.y - y*b.x;}//點積double operator *(const Point &b)const{return x*b.x + y*b.y;}//返回長度double len(){return hypot(x,y);//庫函數}//返回長度的平方double len2(){return x*x + y*y;}//返回兩點的距離double distance(Point p){return hypot(x-p.x,y-p.y);}Point operator +(const Point &b)const{return Point(x+b.x,y+b.y);}Point operator *(const double &k)const{return Point(x*k,y*k);}Point operator /(const double &k)const{return Point(x/k,y/k);}//`計算pa 和 pb 的夾角`//`就是求這個點看a,b 所成的夾角`//`測試 LightOJ1203`double rad(Point a,Point b){Point p = *this;return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));}//`化為長度為r的向量`Point trunc(double r){double l = len();if(!sgn(l))return *this;r /= l;return Point(x*r,y*r);}//`逆時針旋轉90度`Point rotleft(){return Point(-y,x);}//`順時針旋轉90度`Point rotright(){return Point(y,-x);}//`繞著p點逆時針旋轉angle`Point rotate(Point p,double angle){Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);} };struct Line{Point s,e;Line(){}Line(Point _s,Point _e){s = _s;e = _e;}bool operator ==(Line v){return (s == v.s)&&(e == v.e);}//`根據一個點和傾斜角angle確定直線,0<=angle<pi`Line(Point p,double angle){s = p;if(sgn(angle-pi/2) == 0){e = (s + Point(0,1));}else{e = (s + Point(1,tan(angle)));}}//ax+by+c=0Line(double a,double b,double c){if(sgn(a) == 0){s = Point(0,-c/b);e = Point(1,-c/b);}else if(sgn(b) == 0){s = Point(-c/a,0);e = Point(-c/a,1);}else{s = Point(0,-c/b);e = Point(1,(-c-a)/b);}}void input(){s.input();e.input();}void adjust(){if(e < s)swap(s,e);}//求線段長度double length(){return s.distance(e);}//`返回直線傾斜角 0<=angle<pi`double angle(){double k = atan2(e.y-s.y,e.x-s.x);if(sgn(k) < 0)k += pi;if(sgn(k-pi) == 0)k -= pi;return k;}//`點和直線關系`//`1 在左側`//`2 在右側`//`3 在直線上`int relation(Point p){int c = sgn((p-s)^(e-s));if(c < 0)return 1;else if(c > 0)return 2;else return 3;}// 點在線段上的判斷bool pointonseg(Point p){return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;}//`兩向量平行(對應直線平行或重合)`bool parallel(Line v){return sgn((e-s)^(v.e-v.s)) == 0;}//`兩線段相交判斷`//`2 規范相交`//`1 非規范相交`//`0 不相交`int segcrossseg(Line v){int d1 = sgn((e-s)^(v.s-s));int d2 = sgn((e-s)^(v.e-s));int d3 = sgn((v.e-v.s)^(s-v.s));int d4 = sgn((v.e-v.s)^(e-v.s));if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||(d4==0 && sgn((e-v.s)*(e-v.e))<=0);}//`直線和線段相交判斷`//`-*this line -v seg`//`2 規范相交`//`1 非規范相交`//`0 不相交`int linecrossseg(Line v){int d1 = sgn((e-s)^(v.s-s));int d2 = sgn((e-s)^(v.e-s));if((d1^d2)==-2) return 2;return (d1==0||d2==0);}//`兩直線關系`//`0 平行`//`1 重合`//`2 相交`int linecrossline(Line v){if((*this).parallel(v))return v.relation(s)==3;return 2;}//`求兩直線的交點`//`要保證兩直線不平行或重合`Point crosspoint(Line v){double a1 = (v.e-v.s)^(s-v.s);double a2 = (v.e-v.s)^(e-v.s);return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));}//點到直線的距離double dispointtoline(Point p){return fabs((p-s)^(e-s))/length();}//點到線段的距離double dispointtoseg(Point p){if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)return min(p.distance(s),p.distance(e));return dispointtoline(p);}//`返回線段到線段的距離`//`前提是兩線段不相交，相交距離就是0了`double dissegtoseg(Line v){return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));}//`返回點p在直線上的投影`Point lineprog(Point p){return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );}//`返回點p關于直線的對稱點`Point symmetrypoint(Point p){Point q = lineprog(p);return Point(2*q.x-p.x,2*q.y-p.y);} };struct polygon{int n;Point p[maxp];Line l[maxp];void input(int _n){n = _n;for(int i = 0;i < n;i++)p[i].input();}void add(Point q){p[n++] = q;}void getline(){for(int i = 0;i < n;i++){l[i] = Line(p[i],p[(i+1)%n]);}}struct cmp{Point p;cmp(const Point &p0){p = p0;}bool operator()(const Point &aa,const Point &bb){Point a = aa, b = bb;int d = sgn((a-p)^(b-p));if(d == 0){return sgn(a.distance(p)-b.distance(p)) < 0;}return d > 0;}};//`進行極角排序`//`首先需要找到最左下角的點`//`需要重載號好Point的 < 操作符(min函數要用) `void norm(){Point mi = p[0];for(int i = 1;i < n;i++)mi = min(mi,p[i]);sort(p,p+n,cmp(mi));}//`得到凸包`//`得到的凸包里面的點編號是0$\sim$n-1的`//`兩種凸包的方法`//`注意如果有影響，要特判下所有點共點，或者共線的特殊情況`//`測試 LightOJ1203 LightOJ1239`void getconvex(polygon &convex){sort(p,p+n);convex.n = n;for(int i = 0;i < min(n,2);i++){convex.p[i] = p[i];}if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判if(n <= 2)return;int &top = convex.n;top = 1;for(int i = 2;i < n;i++){while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)top--;convex.p[++top] = p[i];}int temp = top;convex.p[++top] = p[n-2];for(int i = n-3;i >= 0;i--){while(top != temp && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)top--;convex.p[++top] = p[i];}if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判convex.norm();//`原來得到的是順時針的點，排序后逆時針`}//`得到凸包的另外一種方法`//`測試 LightOJ1203 LightOJ1239`void Graham(polygon &convex){norm();int &top = convex.n;top = 0;if(n == 1){top = 1;convex.p[0] = p[0];return;}if(n == 2){top = 2;convex.p[0] = p[0];convex.p[1] = p[1];if(convex.p[0] == convex.p[1])top--;return;}convex.p[0] = p[0];convex.p[1] = p[1];top = 2;for(int i = 2;i < n;i++){while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2])) <= 0 )top--;convex.p[top++] = p[i];}if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判}//`判斷是不是凸的`bool isconvex(){bool s[3];memset(s,false,sizeof(s));for(int i = 0;i < n;i++){int j = (i+1)%n;int k = (j+1)%n;s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;if(s[0] && s[2])return false;}return true;}//`判斷點和任意多邊形的關系`//` 3 點上`//` 2 邊上`//` 1 內部`//` 0 外部`int relationpoint(Point q){for(int i = 0;i < n;i++){if(p[i] == q)return 3;}getline();for(int i = 0;i < n;i++){if(l[i].pointonseg(q))return 2;}int cnt = 0;for(int i = 0;i < n;i++){int j = (i+1)%n;int k = sgn((q-p[j])^(p[i]-p[j]));int u = sgn(p[i].y-q.y);int v = sgn(p[j].y-q.y);if(k > 0 && u < 0 && v >= 0)cnt++;if(k < 0 && v < 0 && u >= 0)cnt--;}return cnt != 0;}//`直線u切割凸多邊形左側`//`注意直線方向`//`測試：HDU3982`void convexcut(Line u,polygon &po){int &top = po.n;//注意引用top = 0;for(int i = 0;i < n;i++){int d1 = sgn((u.e-u.s)^(p[i]-u.s));int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));if(d1 >= 0)po.p[top++] = p[i];if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i+1)%n]));}}//`得到周長`//`測試 LightOJ1239`double getcircumference(){double sum = 0;for(int i = 0;i < n;i++){sum += p[i].distance(p[(i+1)%n]);}return sum;}//`得到面積`double getarea(){double sum = 0;for(int i = 0;i < n;i++){sum += (p[i]^p[(i+1)%n]);}return fabs(sum)/2;}//`得到方向`//` 1 表示逆時針，0表示順時針`bool getdir(){double sum = 0;for(int i = 0;i < n;i++)sum += (p[i]^p[(i+1)%n]);if(sgn(sum) > 0)return 1;return 0;}//`得到重心`Point getbarycentre(){Point ret(0,0);double area = 0;for(int i = 1;i < n-1;i++){double tmp = (p[i]-p[0])^(p[i+1]-p[0]);if(sgn(tmp) == 0)continue;area += tmp;ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;}if(sgn(area)) ret = ret/area;return ret;} }Convex;int main() { // freopen("input.txt","r",stdin); // ios::sync_with_stdio(false);int n;while(scanf("%d",&n)!=EOF&&n>=3){double r;scanf("%lf",&r);Point circle;circle.input();Convex.input(n);if(!Convex.isconvex()){puts("HOLE IS ILL-FORMED");continue;}if(Convex.relationpoint(circle)!=1){puts("PEG WILL NOT FIT");continue;}bool flag=true;for(int i=0;i<n;i++){if(sgn(Line(Convex.p[i],Convex.p[(i+1)%n]).dispointtoseg(circle)-r)<0){flag=false;break;}}if(flag)puts("PEG WILL FIT");elseputs("PEG WILL NOT FIT");} return 0; }

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