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 }
zoj2589Circles（平面图的欧拉定理）
时间： 2024-08-25 07:27:04