2017-2018 ACM-ICPC Southeast Regional Contest (Div. 1) F.Move Away 几何

http://codeforces.com/gym/101617/attachments

题意：这个Tom要搬出家里，他想离父母越远越好，但是他有n个朋友，他搬新家的位置要保证离这些朋友不超过他们之间最远的距离。问最后离父母家最多有多远。

输入：第一行一个n，下面n行，朋友的坐标，离第i个朋友最远的距离。

输出离父母（0，0）的距离，保留3位小数。

做法：圆与圆之间的交点，一定是可能的坐标，因为要满足点在所有圆里面，那么这个点至少是两个圆之间的交点，另外对于一个圆最远的距离是圆心+半径。

那么我需要处理两两圆相交的情况的所有交点，判他们是不是在所有圆范围里面，如果是就更新答案。

/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
#define sz size()
typedef long long LL;
typedef pair <int, int> ii;
const int INF=~0U>>1;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=2e4+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=19260817;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

//void readString(string &s)
//{
// static char str[maxn];
// scanf("%s", str);
// s = str;
//}

int n;

int dcmp(db x)
{
if(fabs(x)<eps) return 0;
else return x<0?-1:1;
}
struct Point
{
db x,y;
Point(db x=0,db y=0):x(x),y(y){ }
Point operator + (Point q){ return Point(x+q.x,y+q.y);}
Point operator - (Point q){ return Point(x-q.x,y-q.y);}
Point operator * (double q){ return Point(x*q,y*q);}
bool operator ==(const Point &e)const
{
return dcmp(x-e.x)==0&&dcmp(y-e.y)==0;
}
db len2()
{
return x*x+y*y;
}
db len()
{
return sqrt(x*x+y*y);
}
}Q[maxn];
typedef Point Vector;
db Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
db Length(Vector A){return sqrt(Dot(A,A));}
db angle(Vector v){return atan2(v.y,v.x);}
db Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

struct Line
{
Point P; //直线上一点
Vector v; //方向向量(半平面交中该向量左侧表示相应的半平面)
double ang; //极角，即从x正半轴旋转到向量v所需要的角（弧度）
Line() { } //构造函数
Line(const Line& L): P(L.P), v(L.v), ang(L.ang) { }
Line(const Point& P, const Vector& v): P(P), v(v) { ang = atan2(v.y, v.x); }

bool operator < (const Line& L) const//极角排序
{
return ang < L.ang;
}
Point point(double t) { return P + v*t; }
};

struct Circle
{
Point c;
double r;

Circle(){}
Circle(Point c, double r):c(c), r(r){}
inline void input()
{
scanf("%lf%lf%lf",&c.x,&c.y,&r);
}
Point point(double a)
{
return Point(c.x + r * cos(a), c.y + r * sin(a));
}
}C[maxn];

double DistanceToLine(Point P, Point A, Point B)
{
Vector v1 = B - A, v2 = P - A;
return fabs(Cross(v1, v2) / Length(v1)); //不取绝对值，得到的是有向距离
}
int getLineCircleIntersecion(Line l, Circle C, double& t1, double& t2, vector<Point>& sol)
{
double a = l.v.x;
double b = l.P.x - C.c.x;
double c = l.v.y;
double d = l.P.y - C.c.y;
double e = a * a + c * c;
double f = 2 * (a * b + c * d);
double g = b * b + d * d - C.r * C.r;
double delta = f * f - 4 * e * g;
double dist = DistanceToLine(C.c, l.P, l.P+l.v);
if(dcmp(dist - C.r) == 0)
{ //相切，此处需特殊判断，不能用delta
t1 = t2 = -f / (2 * e);
sol.push_back(l.point(t1));
return 1;
}
if(dcmp(delta) < 0) return 0; //相离
else
{ //相交
t1 = (-f - sqrt(delta)) / (2 * e); sol.push_back(l.point(t1));
t2 = (-f + sqrt(delta)) / (2 * e); sol.push_back(l.point(t2));
return 2;
}
}

int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol)
{
db 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;
db a=angle(C2.c-C1.c);
db 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);
sol.pb(p1);
if(p1==p2) return 1;
sol.pb(p2);
return 2;
}

db dis(Point a,Point b)
{
return (a-b).len();
}
bool check(Point p)
{
for(int i=1;i<=n;i++)
if(dcmp(dis(p,C[i].c)-C[i].r)>0) return 0;
return 1;
}
vector<Point>v;
void solve()
{
s_1(n);
db minc=INF;
FOR(1,n,i)
{
C[i].input();
db t1,t2;
if(C[i].c==(Point){0.0,0.0}){minc=min(minc,C[i].r);n--,i--;}
else getLineCircleIntersecion(Line(C[i].c,C[i].c),C[i],t1,t2,v);
}
if(n==0)
{
printf("%.3f\n",C[1].c.len()+C[1].r);
return ;
}
FOR(1,n,i)
FOR(i+1,n,j)
int xx=getCircleCircleIntersect
4000
ion(C[i],C[j],v);
db ans=0;
// cout<<minc<<endl;
for(int i=0;i<v.size();i++)
{
if(check(v[i])) ans=max(ans,v[i].len());
ans=min(ans,minc);
}
printf("%.3f\n",ans);
}
int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d:\n",cas);
solve();
}
}