Description
Firecrackers scare Nian the monster, but they’re wayyyyy too noisy! Maybe fireworks make a nice complement.
Little Tommy is watching a firework show. As circular shapes spread across the sky, a splendid view unfolds on the night of Lunar New Year’s eve.
A wonder strikes Tommy. How many regions are formed by the circles on the sky? We consider the sky as a flat plane. A region is a connected part of the plane with positive area, whose bound consists of parts of bounds of the circles and is a curve or several curves without self-intersections, and that does not contain any curve other than its boundaries. Note that exactly one of the regions extends infinitely.
Input
The first line of input contains one integer n (1 ≤ n ≤ 3), denoting the number of circles.
The following n lines each contains three space-separated integers x, y and r ( - 10 ≤ x, y ≤ 10, 1 ≤ r ≤ 10), describing a circle whose center is (x, y) and the radius is r. No two circles have the same x, y and r at the same time.
Output
Print a single integer — the number of regions on the plane.
Examples input
3
0 0 1
2 0 1
4 0 1
Examples output
4
题意
给定平面内 n 个圆的信息，求这些圆把平面分成了几个区域。
思路
久违的模板题佯~ 好开心~ 然而模板放在学校了哭唧唧 〒▽〒
求圆拆分平面有多少个区域怎么能离得开平面图的欧拉公式呢？
一般平面图欧拉公式：$f=e-v+c+1$
其中 $e$ 代表边的数量，$v$ 代表点的数量，$c$ 代表连通块的数量，$f$ 代表平面区域的个数
我们把圆弧看作边，交点看作顶点，于是很容易便可以算出 $e,v,c$ 啦~
对于 $e$ ，它相当于每个圆上交点的数量和（因为这些交点把圆拆分成了这么多的弧）
对于 $v$ ，枚举求出交点去重即可
对于 $c$ ，我们已经有了无向图的边，那连通块的数量可以直接 dfs/bfs 或者并查集算出来啦~
然后套用公式就是结果了，注意精度问题。
AC 代码
#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);
using namespace std;
typedef __int64 LL;
typedef long double Ldb;
const int maxn = 10;
const Ldb eps = 1e-12;
int n;
struct Point
{
Ldb x,y;
Point() {}
Point(Ldb _x,Ldb _y):x(_x),y(_y) {}
Point operator + (const Point &t)const
{
return Point(x+t.x,y+t.y);
}
Point operator - (const Point &t)const
{
return Point(x-t.x,y-t.y);
}
Point operator * (const Ldb &t)const
{
return Point(x*t,y*t);
}
Point operator / (const Ldb &t)const
{
return Point(x/t,y/t);
}
bool operator < (const Point &t)const
{
return fabs(x-t.x) < eps ? y<t.y : x<t.x;
}
bool operator == (const Point &t)const
{
return fabs(x-t.x)<eps && fabs(y-t.y)<eps;
}
Ldb len()const
{
return sqrt(x*x+y*y);
}
Point rot90()const
{
return Point(-y,x);
}
};
struct Circle
{
Point o;
Ldb r;
Circle() {}
Circle(Point _o,Ldb _r):o(_o),r(_r) {}
friend vector<Point> operator & (const Circle &c1,const Circle &c2)
{
Ldb d=(c1.o-c2.o).len();
if(d>c1.r+c2.r+eps || d<fabs(c1.r-c2.r)-eps)
return vector<Point>();
Ldb dt=(c1.r*c1.r-c2.r*c2.r)/d,d1=(d+dt)/2;
Point dir=(c2.o-c1.o)/d,pcrs=c1.o+dir*d1;
dt=sqrt(max(0.0L,c1.r*c1.r-d1*d1)),dir=dir.rot90();
return vector<Point> {pcrs+dir*dt,pcrs-dir*dt};
}
} p[maxn];
bool vis[maxn];
int fa[maxn],rk[maxn];
void init()
{
for(int i=1; i<maxn; i++)
fa[i] = i,rk[i] = 0;
}
int find_set(int x)
{
if(fa[x]!=x)
fa[x] = find_set(fa[x]);
return fa[x];
}
void union_set(int x,int y)
{
x = find_set(x);
y = find_set(y);
if(rk[x]>rk[y])
fa[y] = x;
else
{
fa[x] = y;
if(rk[x]==rk[y])
rk[y]++;
}
}
int main()
{
IO;
init();
cin>>n;
for(int i=1; i<=n; i++)
cin>>p[i].o.x>>p[i].o.y>>p[i].r;
int e = 0,v = 0;
vector<Point> all;
for(int i=1; i<=n; i++)
{
vector<Point> tmp1;
for(int j=1; j<=n; j++)
{
if(i==j)
continue;
vector<Point> tmp2 = p[i] & p[j];
if(tmp2.size())
union_set(i,j);
tmp1.insert(tmp1.end(),tmp2.begin(),tmp2.end());
all.insert(all.end(),tmp2.begin(),tmp2.end());
}
sort(tmp1.begin(),tmp1.end());
e += unique(tmp1.begin(),tmp1.end()) - tmp1.begin();
}
sort(all.begin(),all.end());
v = unique(all.begin(),all.end()) - all.begin();
set<int> c;
for(int i=1; i<=n; i++)
c.insert(find_set(i));
cout<<e-v+c.size()+1<<endl;
return 0;
}
