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）
  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 55
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-10;
 15 const double pi = acos(-1.0);
 16 const double inf = ~0u>>2;
 17 
 18 struct point
 19 {
 20     double x,y;
 21     point(double x=0,double y=0):x(x),y(y){}
 22 };
 23 vector<point>ed[N];
 24 vector<int>dd[N];
 25 vector<point>td[N];
 26 struct circle
 27 {
 28     point c;
 29     double r;
 30     point ppoint(double a)
 31     {
 32         return point(c.x+cos(a)*r,c.y+sin(a)*r);
 33     }
 34 };
 35 circle cp[N],cq[N];
 36 int fa[N];
 37 typedef point pointt;
 38 pointt operator -(point a,point b)
 39 {
 40     return point(a.x-b.x,a.y-b.y);
 41 }
 42 
 43 int dcmp(double x)
 44 {
 45     if(fabs(x)<eps) return 0;
 46     return x<0?-1:1;
 47 }
 48 
 49 bool operator == (const point &a,const point &b)
 50 {
 51     return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
 52 }
 53 double dis(point a)
 54 {
 55     return sqrt(a.x*a.x+a.y*a.y);
 56 }
 57 double angle(point a)//计算向量极角
 58 {
 59     return atan2(a.y,a.x);
 60 }
 61 double sqr(double x) {
 62 
 63     return x * x;
 64 
 65 }
 66 
 67 bool intersection(const point& o1, double r1, const point& o2, double r2,int k,int kk)
 68 {
 69     double d = dis(o1- o2);
 70     if (d < fabs(r1 - r2) - eps || d > r1 + r2 + eps)
 71     {
 72         return false;
 73     }
 74     double cosa = (sqr(r1) + sqr(d) - sqr(r2)) / (2 * r1 * d);
 75     double sina = sqrt(max(0., 1. - sqr(cosa)));
 76     point p1 = o1,p2 = o1;
 77     p1.x += r1 / d * ((o2.x - o1.x) * cosa + (o2.y - o1.y) * -sina);
 78     p1.y += r1 / d * ((o2.x - o1.x) * sina + (o2.y - o1.y) * cosa);
 79     p2.x += r1 / d * ((o2.x - o1.x) * cosa + (o2.y - o1.y) * sina);
 80     p2.y += r1 / d * ((o2.x - o1.x) * -sina + (o2.y - o1.y) * cosa);
 81     //cout<<p1.x<<" --"<<p2.x<<" "<<o1.x<<" "<<o1.y<<" "<<o2.x<<" "<<o2.y<<endl;
 82     //printf("%.10f %.10f %.10f %.10f\n",p1.x,p1.y,p2.x,p2.y);
 83     ed[k].push_back(p1);
 84     ed[k].push_back(p2);
 85     ed[kk].push_back(p1);
 86     ed[kk].push_back(p2);
 87 
 88     return true;
 89 }
 90 bool cmp(circle a, circle b)
 91 {
 92     if(dcmp(a.r-b.r)==0)
 93     {
 94         if(dcmp(a.c.x-b.c.x)==0)
 95         return a.c.y<b.c.y;
 96         return a.c.x<b.c.x;
 97     }
 98     return a.r<b.r;
 99 }
100 bool cmpp(point a,point b)
101 {
102     if(dcmp(a.x-b.x)==0)
103     return a.y<b.y;
104     return a.x<b.x;
105 }
106 int find(int x)
107 {
108     if(fa[x]!=x)
109     {
110         fa[x] = find(fa[x]);
111         return fa[x];
112     }
113     return x;
114 }
115 int main()
116 {
117     int t,i,j,n;
118     cin>>t;
119     while(t--)
120     {
121         scanf("%d",&n);
122         for(i = 1; i <= n ;i++)
123         {
124             ed[i].clear();fa[i] = i;
125             dd[i].clear();
126             td[i].clear();
127         }
128         for(i = 1; i <= n ;i++)
129         scanf("%lf%lf%lf",&cp[i].c.x,&cp[i].c.y,&cp[i].r);
130         sort(cp+1,cp+n+1,cmp);
131         int g = 1;
132         cq[g] = cp[g];
133         for(i = 2; i <= n; i++)
134         {
135             if(cp[i].c==cp[i-1].c&&dcmp(cp[i].r-cp[i-1].r)==0)
136             continue;
137             cq[++g] = cp[i];
138         }
139         for(i = 1; i <= g; i++)
140         {
141             for(j = i+1 ; j <= g; j++)
142             {
143                 int flag = intersection(cq[i].c,cq[i].r,cq[j].c,cq[j].r,i,j);
144                 if(flag == 0) continue;
145                 int tx = find(i),ty = find(j);
146                 fa[tx] = ty;
147             }
148         }
149         for(i = 1; i <= g;  i++)
150         {
151             int fx = find(i);
152             dd[fx].push_back(i);
153         }
154         int B=0,V=0;
155         int ans = 0;
156         int num = 0;
157         for(i = 1 ; i <= g; i++)
158         {
159             if(dd[i].size()==0) continue;
160             B = 0,V = 0;
161             for(j = 0 ;j < dd[i].size() ; j++)
162             {
163                 int u = dd[i][j];
164                 if(ed[u].size()==0)
165                 {
166                     B++;
167                     continue;
168                 }
169                 sort(ed[u].begin(),ed[u].end(),cmpp);
170                 td[i].push_back(ed[u][0]);
171                 int o = 1;
172                 for(int e = 1 ; e < ed[u].size() ; e++)
173                 {
174                     //printf("%.10f %.10f\n",ed[u][e].x,ed[u][e].y);
175                     td[i].push_back(ed[u][e]);
176                     if(ed[u][e]==ed[u][e-1]) continue;
177                     else o++;
178                 }
179                 B+=o;
180             }
181             sort(td[i].begin(),td[i].end(),cmpp);
182             V+=1;
183            // cout<<td[i].size()<<endl;
184             for(j = 1; j < td[i].size() ; j++)
185             {
186                 //printf("%.10f %.10f\n",td[i][j].x,td[i][j].y);
187                 if(td[i][j]==td[i][j-1]) continue;
188                 else V++;
189 
190             }
191            // cout<<B+1-V<<" "<<B<<" "<<V<<endl;
192             ans+=B+1-V;
193         }
194         cout<<1+ans<<endl;
195     }
196     return 0;
197 }
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posted @ 2014-08-08 15:31  _雨  阅读(572)  评论(0)  编辑  收藏  举报