zoj2589Circles（平面图的欧拉定理）

链接

连通图中：

设一个平面图形的顶点数为n，划分区域数为r，一笔画笔数为也就是边数m,则有：

n+r-m=2

那么不算外面的那个大区域的话 就可以写为 n+r-m = 1

那么这个题就可以依次求出每个连通图的r = m-n+1 累加起来 最后加上最外面那个平面。

 

注意交点的去重，对于一个圆的边数其实就是交点的数量（排除没有交点的情况）

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 55
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;

 struct point
 {
     double x,y;
     point(double x=,double y=):x(x),y(y){}
 };
 vector<point>ed[N];
 vector<int>dd[N];
 vector<point>td[N];
 struct circle
 {
     point c;
     double r;
     point ppoint(double a)
     {
         return point(c.x+cos(a)*r,c.y+sin(a)*r);
     }
 };
 circle cp[N],cq[N];
 int fa[N];
 typedef point pointt;
 pointt operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }

 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }

 bool operator == (const point &a,const point &b)
 {
     return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;
 }
 double dis(point a)
 {
     return sqrt(a.x*a.x+a.y*a.y);
 }
 double angle(point a)//计算向量极角
 {
     return atan2(a.y,a.x);
 }
 double sqr(double x) {

     return x * x;

 }

 bool intersection(const point& o1, double r1, const point& o2, double r2,int k,int kk)
 {
     double d = dis(o1- o2);
     if (d < fabs(r1 - r2) - eps || d > r1 + r2 + eps)
     {
         return false;
     }
     double cosa = (sqr(r1) + sqr(d) - sqr(r2)) / ( * r1 * d);
     double sina = sqrt(max(., . - sqr(cosa)));
     point p1 = o1,p2 = o1;
     p1.x += r1 / d * ((o2.x - o1.x) * cosa + (o2.y - o1.y) * -sina);
     p1.y += r1 / d * ((o2.x - o1.x) * sina + (o2.y - o1.y) * cosa);
     p2.x += r1 / d * ((o2.x - o1.x) * cosa + (o2.y - o1.y) * sina);
     p2.y += r1 / d * ((o2.x - o1.x) * -sina + (o2.y - o1.y) * cosa);
     //cout<<p1.x<<" --"<<p2.x<<" "<<o1.x<<" "<<o1.y<<" "<<o2.x<<" "<<o2.y<<endl;
     //printf("%.10f %.10f %.10f %.10f\n",p1.x,p1.y,p2.x,p2.y);
     ed[k].push_back(p1);
     ed[k].push_back(p2);
     ed[kk].push_back(p1);
     ed[kk].push_back(p2);

     return true;
 }
 bool cmp(circle a, circle b)
 {
     if(dcmp(a.r-b.r)==)
     {
         if(dcmp(a.c.x-b.c.x)==)
         return a.c.y<b.c.y;
         return a.c.x<b.c.x;
     }
     return a.r<b.r;
 }
 bool cmpp(point a,point b)
 {
     if(dcmp(a.x-b.x)==)
     return a.y<b.y;
     return a.x<b.x;
 }
 int find(int x)
 {
     if(fa[x]!=x)
     {
         fa[x] = find(fa[x]);
         return fa[x];
     }
     return x;
 }
 int main()
 {
     int t,i,j,n;
     cin>>t;
     while(t--)
     {
         scanf("%d",&n);
         for(i = ; i <= n ;i++)
         {
             ed[i].clear();fa[i] = i;
             dd[i].clear();
             td[i].clear();
         }
         for(i = ; i <= n ;i++)
         scanf("%lf%lf%lf",&cp[i].c.x,&cp[i].c.y,&cp[i].r);
         sort(cp+,cp+n+,cmp);
         int g = ;
         cq[g] = cp[g];
         for(i = ; i <= n; i++)
         {
             if(cp[i].c==cp[i-].c&&dcmp(cp[i].r-cp[i-].r)==)
             continue;
             cq[++g] = cp[i];
         }
         for(i = ; i <= g; i++)
         {
             for(j = i+ ; j <= g; j++)
             {
                 int flag = intersection(cq[i].c,cq[i].r,cq[j].c,cq[j].r,i,j);
                 if(flag == ) continue;
                 int tx = find(i),ty = find(j);
                 fa[tx] = ty;
             }
         }
         for(i = ; i <= g;  i++)
         {
             int fx = find(i);
             dd[fx].push_back(i);
         }
         int B=,V=;
         int ans = ;
         int num = ;
         for(i =  ; i <= g; i++)
         {
             if(dd[i].size()==) continue;
             B = ,V = ;
             for(j =  ;j < dd[i].size() ; j++)
             {
                 int u = dd[i][j];
                 if(ed[u].size()==)
                 {
                     B++;
                     continue;
                 }
                 sort(ed[u].begin(),ed[u].end(),cmpp);
                 td[i].push_back(ed[u][]);
                 int o = ;
                 for(int e =  ; e < ed[u].size() ; e++)
                 {
                     //printf("%.10f %.10f\n",ed[u][e].x,ed[u][e].y);
                     td[i].push_back(ed[u][e]);
                     if(ed[u][e]==ed[u][e-]) continue;
                     else o++;
                 }
                 B+=o;
             }
             sort(td[i].begin(),td[i].end(),cmpp);
             V+=;
            // cout<<td[i].size()<<endl;
             for(j = ; j < td[i].size() ; j++)
             {
                 //printf("%.10f %.10f\n",td[i][j].x,td[i][j].y);
                 if(td[i][j]==td[i][j-]) continue;
                 else V++;

             }
            // cout<<B+1-V<<" "<<B<<" "<<V<<endl;
             ans+=B+-V;
         }
         cout<<+ans<<endl;
     }
     return ;
 }