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-10;
const double pi = acos(-1.0);
const double inf = ~0u>>2;

struct point
{
    double x,y;
    point(double x=0,double y=0):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 0;
    return x<0?-1:1;
}

bool operator == (const point &a,const point &b)
{
    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
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)) / (2 * r1 * d);
    double sina = sqrt(max(0., 1. - 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
",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)==0)
    {
        if(dcmp(a.c.x-b.c.x)==0)
        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)==0)
    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 = 1; i <= n ;i++)
        {
            ed[i].clear();fa[i] = i;
            dd[i].clear();
            td[i].clear();
        }
        for(i = 1; i <= n ;i++)
        scanf("%lf%lf%lf",&cp[i].c.x,&cp[i].c.y,&cp[i].r);
        sort(cp+1,cp+n+1,cmp);
        int g = 1;
        cq[g] = cp[g];
        for(i = 2; i <= n; i++)
        {
            if(cp[i].c==cp[i-1].c&&dcmp(cp[i].r-cp[i-1].r)==0)
            continue;
            cq[++g] = cp[i];
        }
        for(i = 1; i <= g; i++)
        {
            for(j = i+1 ; j <= g; j++)
            {
                int flag = intersection(cq[i].c,cq[i].r,cq[j].c,cq[j].r,i,j);
                if(flag == 0) continue;
                int tx = find(i),ty = find(j);
                fa[tx] = ty;
            }
        }
        for(i = 1; i <= g;  i++)
        {
            int fx = find(i);
            dd[fx].push_back(i);
        }
        int B=0,V=0;
        int ans = 0;
        int num = 0;
        for(i = 1 ; i <= g; i++)
        {
            if(dd[i].size()==0) continue;
            B = 0,V = 0;
            for(j = 0 ;j < dd[i].size() ; j++)
            {
                int u = dd[i][j];
                if(ed[u].size()==0)
                {
                    B++;
                    continue;
                }
                sort(ed[u].begin(),ed[u].end(),cmpp);
                td[i].push_back(ed[u][0]);
                int o = 1;
                for(int e = 1 ; e < ed[u].size() ; e++)
                {
                    //printf("%.10f %.10f
",ed[u][e].x,ed[u][e].y);
                    td[i].push_back(ed[u][e]);
                    if(ed[u][e]==ed[u][e-1]) continue;
                    else o++;
                }
                B+=o;
            }
            sort(td[i].begin(),td[i].end(),cmpp);
            V+=1;
           // cout<<td[i].size()<<endl;
            for(j = 1; j < td[i].size() ; j++)
            {
                //printf("%.10f %.10f
",td[i][j].x,td[i][j].y);
                if(td[i][j]==td[i][j-1]) continue;
                else V++;

            }
           // cout<<B+1-V<<" "<<B<<" "<<V<<endl;
            ans+=B+1-V;
        }
        cout<<1+ans<<endl;
    }
    return 0;
}