P1355 神秘大三角（凸包）
P1355 神秘大三角 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
队友推荐的，算是入门凸包，就是用叉积判断一下点是否相对每条边都在凸包的边的左侧。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define ll long long 6 7 const int N=1e3+10; 8 9 double eps=1e-8; 10 11 double pi=acos(-1); 12 13 struct Point{ 14 15 double x,y; 16 17 Point(double x=0,double y=0):x(x),y(y){}; 18 19 }; 20 21 typedef Point Vector; 22 23 Vector operator +(Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);} 24 25 Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);} 26 27 Vector operator *(Vector A,double B){return Vector(A.x*B,A.y*B);} 28 29 Vector operator /(Vector A,double B){return Vector(A.x/B,A.y/B);} 30 31 int dcmp(double x){ 32 33 if(fabs(x)<eps)return 0; 34 35 return x<0?-1:1; 36 37 } 38 39 bool operator<(const Point&a,const Point&b){return a.x<b.x||(a.x==b.x&&a.y<b.y);} 40 41 bool operator == (const Point &a,const Point &b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;} 42 43 double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;} 44 45 double Length(Vector A){return sqrt(Dot(A,A));} 46 47 double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;} 48 49 double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));} 50 51 double Angle(Vector A) {return atan2(A.y, A.x);} 52 53 54 55 struct Circle{ //圆 56 57 Point c; 58 59 double r; 60 61 Circle(Point c = Point(0, 0), double r = 0):c(c),r(r){} 62 63 Point point(double a){ 64 65 return Point(c.x+cos(a)*r,c.y+sin(a)*r); 66 67 } 68 69 }; 70 71 double dis(Point A,Point B){ 72 73 return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)); 74 75 } 76 77 int circle_circle(Circle C1,Circle C2,vector<Point>&sol){//返回圆与圆的交点个数 78 79 double d=Length(C1.c-C2.c); 80 81 if(dcmp(d)==0){ 82 83 if(dcmp(C1.r-C2.r)==0)return -1; 84 85 return 0; 86 87 } 88 89 if(dcmp(C1.r+C2.r-d)<0)return 0; 90 91 if(dcmp(fabs(C1.r-C2.r)-d)>0)return 0; 92 93 double a=Angle(C2.c-C1.c); 94 95 double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*d*C1.r)); 96 97 Point p1=C1.point(a-da),p2=C1.point(a+da); 98 99 sol.push_back(p1); 100 101 if(p1==p2)return 1; 102 103 sol.push_back(p2); 104 105 return 2; 106 107 } 108 109 110 111 Point line_point(Point p,Vector v,Point q,Vector w){//直线交点 112 113 Vector u=p-q; 114 115 double t=Cross(w,u)/Cross(v,w); 116 117 return p+v*t; 118 119 } 120 121 bool onsegment(Point p,Point a1,Point a2){ 122 123 if(p==a1||p==a2)return true; 124 125 return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0; 126 127 } 128 129 bool segmentcross(Point a1,Point a2,Point b1,Point b2){ 130 131 double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1), 132 133 c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1); 134 135 return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; 136 137 } 138 139 Point line_line(Point A,Point B,Point C,Vector D){//线段交点 140 141 if(segmentcross(A,B,C,D)){ 142 143 return line_point(A,B-A,C,D-C); 144 145 } 146 147 return Point(123.456,0); 148 149 } 150 151 int tubao(Point *p,int n,Point *ch){ 152 153 sort(p,p+n); 154 155 n=unique(p,p+n)-p; 156 157 int m=0; 158 159 for(int i=0;i<n;i++){ 160 161 while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--; 162 163 ch[m++]=p[i]; 164 165 } 166 167 int k=m; 168 169 for(int i=n-2;i>=0;i--){ 170 171 while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--; 172 173 ch[m++]=p[i]; 174 175 } 176 177 if(n>1)m--; 178 179 return m; 180 181 } 182 183 double tubaos(Point *p,int n){ 184 185 double area=0; 186 187 for(int i=1;i<n-1;i++){ 188 189 area+=Cross(p[i]-p[0],p[i+1]-p[0]); 190 191 } 192 193 return area/2; 194 195 } 196 197 int intubao(Point *ch,int n,Point p){ 198 199 Vector A,B; 200 201 int flag=0; 202 203 for(int i=0;i<n;i++){ 204 if(ch[i] == p)return -2; //p在凸包端点上 205 206 A=ch[(i+1)%n]-ch[i]; 207 208 B=p-ch[i]; 209 210 if(onsegment(p,ch[i],ch[(i+1)%n])){ 211 212 flag=-1;//在凸包边上 213 214 break; 215 216 } 217 218 else if(Cross(A,B)>0){ 219 220 flag++; 221 222 } 223 224 } 225 226 if(flag==-1||flag==n)return flag;//flag=n说明在凸包里面 227 228 return 0;//在凸包外面 229 230 } 231 232 double x[N],y[N]; 233 234 Point p[N],q[N],ch[N],ans[N]; 235 236 int main(){ 237 238 char c; 239 240 for(int i = 0;i < 4;i++){ 241 cin >> c >> p[i].x >> c >> p[i].y >> c; 242 } 243 244 int n = tubao(p, 3, ch); 245 int res = intubao(ch, n, p[3]); 246 if(res == 3)cout << 1 << endl; 247 else if(res == 0)cout << 2 << endl; 248 else if(res == -1)cout << 3 << endl; 249 else cout << 4 << endl; 250 251 252 }
