POJ 1418 圆的基本操作以及 圆弧离散化

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Description

Do you know confetti? They are small discs of colored paper, and people throw them around during parties or festivals. Since people throw lots of confetti, they may end up stacked one on another, so there may be hidden ones underneath.

A handful of various sized confetti have been dropped on a table. Given their positions and sizes, can you tell us how many of them you can see?

The following figure represents the disc configuration for the first sample input, where the bottom disc is still visible.

Input

Output

Sample Input

3
0 0 0.5
-0.9 0 1.00000000001
0.9 0 1.00000000001
5
0 1 0.5
1 1 1.00000000001
0 2 1.00000000001
-1 1 1.00000000001
0 -0.00001 1.00000000001
5
0 1 0.5
1 1 1.00000000001
0 2 1.00000000001
-1 1 1.00000000001
0 0 1.00000000001
2
0 0 1.0000001
0 0 1
2
0 0 1
0.00000001 0 1
0

Sample Output

3
5
4
2
2

给定一堆圆，求可见的圆有几个。

问别人的思路；

/* ***********************************************
Author :rabbit
Created Time :2014/7/8 13:49:36
File Name :3.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-14
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
	if(fabs(x)<eps)return 0;
	return x>0?1:-1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0){
		x=_x;y=_y;
	}
};
Point operator + (Point a,Point b){
	return Point(a.x+b.x,a.y+b.y);
}
Point operator - (Point a,Point b){
	return Point(a.x-b.x,a.y-b.y);
}
Point operator * (Point a,double p){
	return Point(a.x*p,a.y*p);
}
Point operator / (Point a,double p){
	return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
	return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
	return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Length(Point a){
	return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
	return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
	return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
	return a/Length(a);
}
Point Normal(Point a){
	return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
	return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
	Point u=p-q;
	double t=Cross(w,u)/Cross(v,w);
	return p+v*t;
}
struct Line{
	Point p,v;
	double ang;
	Line(){}
	Line(Point _p,Point _v):p(_p),v(_v){
		ang=atan2(v.y,v.x);
	}
	Point point(double a){
		return p+(v*a);
	}
	bool operator < (const Line &L) const{
		return ang<L.ang;
	}
};
Point GetLineIntersection(Line a,Line b){
	return GetLineIntersection(a.p,a.v,b.p,b.v);
}
struct Circle{
	Point c;
	double r;
	Circle(){}
	Circle(Point _c,double _r):c(_c),r(_r){}
	Point point(double a){
		return Point(c.x+cos(a)*r,c.y+sin(a)*r);
	}
};
Circle C[200];
bool vis[200];
vector<double> pp[200];
int GetCircleCircleIntersection(int s1,int s2){
	Circle c1=C[s1],c2=C[s2];
	double d=Length(c1.c-c2.c);
	if(dcmp(d)==0){
		if(dcmp(c1.r-c2.r)==0)return -1;
		return 0;
	}
	if(dcmp(c1.r+c2.r-d)<0)return 0;
	if(dcmp(fabs(c1.r-c2.r)-d)>0)return 0;
	double a=angle(c2.c-c1.c);
	double da=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
	Point p1=c1.point(a-da),p2=c1.point(a+da);
	if(p1==p2)return 1;
	pp[s1].push_back(a+da);
	pp[s1].push_back(a-da);
	return 2;
}
bool PointInCircle(Point p, Circle C){      
    double dist = Length(p - C.c);
    if(dcmp(dist - C.r) > 0) return 0;      
    else return 1;
}
bool CircleInCircle(Circle A, Circle B){       
    double cdist = Length(A.c - B.c);
    double rdiff = B.r - A.r;
    if(dcmp(A.r - B.r) <= 0 && dcmp(cdist - rdiff) <= 0) return 1;     
    return 0;
}
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n;
	 while(~scanf("%d",&n)&&n){
		 memset(vis,0,sizeof(vis));
		 for(int i=0;i<n;i++)pp[i].clear();
		 for(int i=0;i<n;i++)
			 scanf("%lf%lf%lf",&C[i].c.x,&C[i].c.y,&C[i].r);
		 for(int i=0;i<n;i++)
			 for(int j=0;j<n;j++){
				 if(i==j)continue;
				 GetCircleCircleIntersection(i,j);
			 }
		 for(int i=0;i<n;i++){
			 sort(pp[i].begin(),pp[i].end());
			 pp[i].resize(unique(pp[i].begin(),pp[i].end())-pp[i].begin());
		 }
		 for(int i=0;i<n;i++){
			 if(pp[i].size()==0){
				 bool ok=1;
				 for(int j=i+1;j<n;j++)
					 if(CircleInCircle(C[i],C[j])){
						 ok=0;break;
					 }
				 if(ok)vis[i]=1;
			//	 cout<<"han->1"<<endl;
			 }
			 else{
			//	 cout<<"han->2"<<endl;
				 int sz=pp[i].size();
				 pp[i].push_back(pp[i][0]);
				 for(int j=0;j<sz;j++){
					 Point dd=C[i].point((pp[i][j]+pp[i][j+1])/2);
					 bool ok=1;
					 for(int k=i+1;k<n;k++)
						 if(PointInCircle(dd,C[k])){
						//	 cout<<dd.x<<" "<<dd.y<<" "<<k<<endl;
							 ok=0;break;
						 }
					 if(ok){
						 vis[i]=1;
						 for(int k=i-1;k>=0;k--)
							 if(PointInCircle(dd,C[k])){
								 vis[k]=1;break;
							 }
					 }
				 }
			 }
		 }
		 int ans=0;
	//	 cout<<"han  ";for(int i=0;i<n;i++)cout<<vis[i]<<" ";cout<<endl;
		 for(int i=0;i<n;i++)
			 if(vis[i])ans++;
		 cout<<ans<<endl;
	 }
     return 0;
}

POJ 1418 圆的基本操作以及 圆弧离散化,布布扣,bubuko.com

POJ 1418 圆的基本操作以及 圆弧离散化

标签：des   style   blog   http   color   strong   

原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/37565971