LA 2572 Viva Confetti 离散化 *

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题目地址：https://vjudge.net/problem/UVALive-2572

离散化：把每个圆看成是一段一段弧组成的

每个圆拿出来看一下，先求其上与其他所有圆的交点。每两个交点之间的那段弧要么可见要么不可见，在弧上任取一个点(这里取的的中点)，将该点向内(圆心方向)和向外移动一个微小的eps距离，为p1，p2，看看p1和p2是否在最上层的圆内，如果在内部，那么该圆可见

#include <bits/stdc++.h>using namespace std;const int maxn=100+5;const double PI=acos(-1);const double EPS=1e-13;#define REP(i,a,b)  for(int i=a;i<=(int)(b);++i)#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)int dcmp(double x){if(x>EPS) return 1;else if(x<-EPS) return -1;else return 0;}struct Point{double x,y;Point(double x=0,double y=0):x(x),y(y){}};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); } Vector operator - (Vector A, Vector B) { return Vector(A.x-B.x, A.y-B.y); } Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }double Angle(Vector v) { return atan2(v.y,v.x); }double Length(Vector v) { return sqrt(v.x*v.x+v.y*v.y); }double NormalizeAngle(double a){   //把(-180,180] 转化为 [0,360)return fmod(a+2*PI,2*PI);}void getCircleCircleIntersection(Point C1, double r1, Point C2, double r2,vector<double>& sol){double d = Length(C1-C2);if(dcmp(d)==0) return;if(dcmp(r1+r2-d)<0) return;if(dcmp(fabs(r1-r2)-d>0)) return; //内含，相离double ang=Angle(C2-C1);double dang=acos((r1*r1+d*d-r2*r2)/2/r1/d);sol.push_back(NormalizeAngle(ang+dang));if(dcmp(dang)!=0) sol.push_back(NormalizeAngle(ang-dang));}Point C[maxn*2];double RC[maxn];int n;bool vis[maxn];vector<double> rad;inline Point point(Point p,double ang,double r){return Point(p.x+r*cos(ang),p.y+r*sin(ang));}void UpdateTop(Point p){REPD(i,n,1){double dist=Length(p-C[i]);if(dcmp(dist-RC[i])<0) { vis[i]=true; break; }}}int main(int argc, char const *argv[]){// freopen("input.in","r",stdin);while(scanf("%d",&n)==1&&n){REP(i,1,n){double x,y,r;scanf("%lf%lf%lf",&x,&y,&r);C[i]=Point(x,y); RC[i]=r;}memset(vis,false,sizeof(vis));REP(i,1,n){rad.clear();rad.push_back(0);rad.push_back(2*PI);REP(j,1,n) {if(j==i) continue;getCircleCircleIntersection(C[i],RC[i],C[j],RC[j],rad);}sort(rad.begin(), rad.end());vector<double>::iterator it=unique(rad.begin(), rad.end());rad.erase(it,rad.end());REP(k,0,rad.size()-2){for(int side=-1;side<=1;side+=2){double nr=RC[i]+side*EPS*2;UpdateTop(point(C[i],(rad[k]+rad[k+1])/2,nr));}}}int cnt=0;REP(i,1,n) if(vis[i]) cnt++;printf("%d\n", cnt);}return 0;}

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