计算几何板子
#include <iostream> #include <string.h> #include <queue> #include <stdio.h> #include <algorithm> #include <math.h> typedef long long ll; using namespace std; const int maxn=1e4+10; const int mod=998244353; using namespace std; const ll eps = 1e-8; int sgn(ll x) { if(fabs(x) < eps)return 0; if(x < 0) return -1; return 1; } struct Point { ll x,y; Point(){} Point(ll _x,ll _y) { x = _x;y = _y; } Point operator -(const Point &b)const { return Point(x - b.x,y - b.y); } ll operator ^(const Point &b)const { return x*b.y - y*b.x; } ll operator *(const Point &b)const { return x*b.x + y*b.y; } ll operator ==(const Point &b)const { return x==b.x&&y==b.y; }
}; struct Line { Point s,e; Line(){} Line(Point _s,Point _e) { s = _s;e = _e; } }; ll xmult(Point p0,Point p1,Point p2) //p0p1 X p0p2 { return (p1-p0)^(p2-p0); } ll xmult(Point p0,Point p1,Point p2,Point p3) //p0p1 X p2p3 { return (p1-p0)^(p3-p2); } bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交 { return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= 0; } ll dist(Point a,Point b) { return sqrt( (b - a)*(b - a) ); } //多边形面积s的二倍(参数按逆时针给出) ll get_mianji(Point *P,int num) { ll ans=0; for(int i=0;i<num;i++) { ans+=P[i]^P[(i+1)%num]; } return ans; } //多边形形面积e
//凸包s Point tubao[maxn]; int num_tubao; Point t(0,1e9+7); bool cmp_jijiao(Point p1,Point p2) { if(xmult(t, p1, p2)>0) { return 1; } else if(xmult(t, p1, p2)==0) { return dist(t, p1)<dist(t, p2); } else return 0; } void get_tubao(Point *P,int num) { t.x=0; t.y=1e9+7; for(int i=0;i<num;i++) { if(P[i].y<t.y) { t=P[i]; } else if(P[i].y==t.y) { t.x=min(P[i].x,t.x); } } sort(P, P+num, cmp_jijiao); num_tubao=2; tubao[0]=P[0];tubao[1]=P[1]; for(int i=2;i<num;i++) { while(xmult(tubao[num_tubao-2], tubao[num_tubao-1], P[i])<0) { num_tubao--; } tubao[num_tubao]=P[i]; num_tubao++; } } //凸包e //旋转卡壳s(求最大四边形面积) ll ro_ka(Point *P,int num) { ll ans=0; //去掉重复的点s int num2=1; for(int i=1;i<num;i++) { if(P[i].x==P[i-1].x&&P[i].y==P[i-1].y) { continue; } else { P[num2]=P[i]; num2++; } } num=num2; //去掉重复的点e for(int i=0;i<num;i++) P[i+num]=P[i]; //检查是否凸包上的点全部共线s int flag=1; for(int i=2;i<num;i++) { if(xmult(P[0], P[1], P[2])!=0) flag=0; } if(flag==1) return 0; //检查是否凸包上的点全部共线e for(int i=0;i<num;i++){ int l=i,r; for(int j=i+1;j<num;j++) { if(xmult(P[l], P[l+1], P[j],P[j+1])<0){ r=j; break; } } for(int j=i+1;j<num;j++) { while(xmult(P[l], P[l+1], P[i],P[j])>0) l++; while(xmult(P[r], P[r+1], P[i],P[j])<0) r++; Point t[4]={P[i],P[l],P[j],P[r]}; ans=max(ans,get_mianji(t, 4)); } } return ans; } //旋转卡壳e
//坐标旋转s Point rotate(Point p,double angle){ Point v = p; double c = cos(angle), s = sin(angle); return Point(v.x*c - v.y*s , v.x*s + v.y*c); } //坐标旋转e
//半平面交s Point dad_anticlock[maxn];//输入弄好就行了 Line dad_half_line[maxn],que[maxn]; //得到极角角度 double getAngle(Point a) { return atan2(a.y, a.x); } //得到极角角度 double getAngle(Line a) { return atan2(a.e.y - a.s.y, a.e.x - a.s.x); } bool onRight(Line a, Line b, Line c) { Point o = getIntersectPoint(b, c); if (((a.e - a.s) ^ (o - a.s)) < 0) return true; return false; } bool cmp_half(Line L1,Line L2) { Point va = L1.e - L1.s, vb = L2.e - L2.s; double A = getAngle(va), B = getAngle(vb); if (fabs((double)(A - B)) < eps) return ((va) ^ (L2.e - L1.s)) >= 0; return A < B; } bool Is_half(int num) { for(int i=0;i<num;i++) { dad_half_line[i].s=dad_anticlock[i]; dad_half_line[i].e=dad_anticlock[(i+1)%num]; } sort(dad_half_line, dad_half_line+num, cmp_half); int head = 0, tail = 0, cnt = 0;//模拟双端队列 //去重,极角相同时取最后一个。 for (int i = 0; i < num - 1; i++) { if (fabs(getAngle(dad_half_line[i]) - getAngle(dad_half_line[i + 1])) < eps) { continue; } dad_half_line[cnt++] = dad_half_line[i]; } dad_half_line[cnt++] = dad_half_line[num - 1]; for (int i = 0; i < cnt; i++) { //判断新加入直线产生的影响 while(tail - head > 1 && onRight(dad_half_line[i], que[tail - 1], que[tail - 2])) tail--; while(tail - head > 1 && onRight(dad_half_line[i], que[head], que[head + 1])) head++; que[tail++] = dad_half_line[i]; } //最后判断最先加入的直线和最后的直线的影响 while(tail - head > 1 && onRight(que[head], que[tail - 1], que[tail - 2])) tail--; while(tail - head > 1 && onRight(que[tail - 1], que[head], que[head + 1])) head++; if (tail - head < 3) return false; return true; } //半平面交e