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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