zoj2589

Posted 2022-06-25 mxang

是cf933C的升级版。
平面图欧拉定理。over！
f=e-v+c+1
c是联通块，相交才视为一块。
e是圆弧数，v是顶点数。

#include <bits/stdc++.h>
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k)
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;

int cmp(db k1,db k2)return sign(k1-k2);
int inmid(db k1,db k2,db k3)return sign(k1-k3)*sign(k2-k3)<=0;// k3 在 [k1,k2] 内
struct point
    db x,y;
    point operator + (const point &k1) constreturn (point)k1.x+x,k1.y+y;
    point operator - (const point &k1) constreturn (point)x-k1.x,y-k1.y;
    point operator * (db k1) constreturn (point)x*k1,y*k1;
    point operator / (db k1) constreturn (point)x/k1,y/k1;
    int operator == (const point &k1) constreturn cmp(x,k1.x)==0&&cmp(y,k1.y)==0;
    // 逆时针旋转
    point turn(db k1)return (point)x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1);
    point turn90()return (point)-y,x;
    bool operator < (const point k1) const
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    
    db abs()return sqrt(x*x+y*y);
    db abs2()return x*x+y*y;
    db dis(point k1)return ((*this)-k1).abs();
    point unit()db w=abs(); return (point)x/w,y/w;
    void scan()double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;
    void print()printf("%.11lf %.11lf\n",x,y);
    db getw()return atan2(y,x);
    point getdel()if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);
    int getP() constreturn sign(y)==1||(sign(y)==0&&sign(x)==-1);
;
int inmid(point k1,point k2,point k3)return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);
db cross(point k1,point k2)return k1.x*k2.y-k1.y*k2.x;
db dot(point k1,point k2)return k1.x*k2.x+k1.y*k2.y;
db rad(point k1,point k2)return atan2(cross(k1,k2),dot(k1,k2));
// -pi -> pi
int compareangle (point k1,point k2)//极角排序+
    return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0);

point proj(point k1,point k2,point q) // q 到直线 k1,k2 的投影
    point k=k2-k1;return k1+k*(dot(q-k1,k)/k.abs2());

point reflect(point k1,point k2,point q)return proj(k1,k2,q)*2-q;
int clockwise(point k1,point k2,point k3)// k1 k2 k3 逆时针 1 顺时针 -1 否则 0
    return sign(cross(k2-k1,k3-k1));

int checkLL(point k1,point k2,point k3,point k4)// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点
    return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=0;

point getLL(point k1,point k2,point k3,point k4)
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);

int intersect(db l1,db r1,db l2,db r2)
    if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;

int checkSS(point k1,point k2,point k3,point k4)
    return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
           sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
           sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;

db disSP(point k1,point k2,point q)
    point k3=proj(k1,k2,q);
    if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));

db disSS(point k1,point k2,point k3,point k4)
    if (checkSS(k1,k2,k3,k4)) return 0;
    else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));

int onS(point k1,point k2,point q)return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;
struct circle
    point o; db r;
    void scan()o.scan(); scanf("%lf",&r);
    int inside(point k)return cmp(r,o.dis(k));
;
int checkposCC(circle k1,circle k2)// 返回两个圆的公切线数量
    if (cmp(k1.r,k2.r)==-1) swap(k1,k2);
    db dis=k1.o.dis(k2.o);  int w1=cmp(dis,k1.r+k2.r),w2=cmp(dis,k1.r-k2.r);
    if (w1>0) return 4; else if (w1==0) return 3; else if (w2>0) return 2;
    else if (w2==0) return 1; else return 0;

vector<point> getCC(circle k1,circle k2)// 沿圆 k1 逆时针给出 , 相切给出两个
    int pd=checkposCC(k1,k2); if (pd==0||pd==4) return ;
    db a=(k2.o-k1.o).abs2(),cosA=(k1.r*k1.r+a-k2.r*k2.r)/(2*k1.r*sqrt(max(a,(db)0.0)));
    db b=k1.r*cosA,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
    point k=(k2.o-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
    return m-del,m+del;

int t,n,vis[55];
circle c[55];
set<point> s;
vector<point> v;
set<pii> st;
void slove(int x)
    vis[x]=1;
    for(auto p:st)
        for(int i=0;i<n;i++)
            if(vis[p.fi]||vis[p.se])
                vis[p.fi]=vis[p.se]=1;
                break;
            
        
    

int main()
    scanf("%d",&t);
    while (t--) 
        v.clear();
        s.clear();
        st.clear();
        memset(vis,0, sizeof(vis));
        scanf("%d", &n);
        for (int i = 0; i < n; i++) 
            scanf("%lf%lf%lf", &c[i].o.x, &c[i].o.y, &c[i].r);
        
        int p = 0;
        for (int i = 0; i < n; i++) 
            for (int j = i + 1; j < n; j++) 
                v = getCC(c[i], c[j]);
                for (auto t:v)s.insert(t);
                if(!v.empty())
                    st.insert(pii(i,j));
            
        
        for(int i=0;i<n;i++)
            if(!vis[i]) 
                slove(i);
                p++;
            
        
        int e=0,v;
        v = s.size();
        for (int i = 0; i < n; i++) 
            for (auto x:s) 
                if (c[i].inside(x) == 0)
                    e++;
            
        
        cout << e - v + 1 + p << endl;