[POJ 2588] Snakes

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Snakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1015		Accepted: 341

Description

Buffalo Bill wishes to cross a 1000x1000 square field. A number of snakes are on the field at various positions, and each snake can strike a particular distance in any direction. Can Bill make the trip without being bitten?

Input

Assume that the southwest corner of the field is at (0,0) and the northwest corner at (0,1000). The input consists of a line containing n <= 1000, the number of snakes. A line follows for each snake, containing three real numbers: the (x,y) location of the snake and its strike distance. The snake will bite anything that passes closer than this distance from its location.

Output

Bill must enter the field somewhere between the southwest and northwest corner and must leave somewhere between the southeast and northeast corners. 

If Bill can complete the trip, give coordinates at which he may enter and leave the field. If Bill may enter and leave at several places, give the most northerly. If there is no such pair of positions, print "Bill will be bitten." 

Sample Input

3

500 500 499

0 0 999

1000 1000 200

Sample Output

Bill enters at (0.00, 1000.00) and leaves at (1000.00, 800.00).

并查集+计算几何

#include<iostream>

#include<algorithm>

#include<cmath>

#include<vector>

#include<cstdio>

using namespace std;



#define PI acos(-1.0)

#define EPS 1e-8

#define N 1010



int dcmp(double x)

{

    if(fabs(x)<EPS) return 0;

    return x<0?-1:1;

}

struct Point

{

    double x,y;

    Point (){}

    Point (double x,double y):x(x),y(y){}

    Point operator - (Point p){

        return Point(x-p.x,y-p.y);

    }

    bool operator == (Point p){

        return dcmp(fabs(x-p.x))==0 && dcmp(fabs(y-p.y))==0;

    }

    double operator * (Point p){

        return x*p.x+y*p.y;

    }

    double operator ^ (Point p){

        return x*p.y-y*p.x;

    }

    double length(){

        return sqrt(x*x+y*y);

    }

    double angle(){

        return atan2(y,x);

    }

    bool operator <(const Point &p)const{

        return y<p.y;

    }

};

struct Line

{

    Point s,e;

    Line (){}

    Line (Point s,Point e):s(s),e(e){}

    Point GetPoint(double t){

        return Point(s.x+(e.x-s.x)*t,s.y+(e.y-s.y)*t);

    }

};

struct Circle

{

    Point c;

    double r;

    Circle(){}

    Circle(Point c,double r):c(c),r(r){}

    Point GetPoint(double a){

        return Point(c.x+cos(a)*r,c.y+sin(a)*r);

    }

    /* 0表示相离，1表示相切，2表示相交 */

    pair<int,vector<Point> > CircleInterLine(Line l){

        vector<Point> res;

        double A=l.e.x-l.s.x,B=l.s.x-c.x,C=l.e.y-l.s.y,D=l.s.y-c.y;

        double E=A*A+C*C,F=2*(A*B+C*D),G=B*B+D*D-r*r;

        double delta=F*F-4*E*G;

        if(dcmp(delta)<0) return make_pair(0,res);

        if(dcmp(delta)==0){

            res.push_back(l.GetPoint(-F/(2*E)));

            return make_pair(1,res);

        }

        res.push_back(l.GetPoint((-F-sqrt(delta))/(2*E)));

        res.push_back(l.GetPoint((-F+sqrt(delta))/(2*E)));

        return make_pair(2,res);

    }

    /* -1表示重合，0表示相离，1表示相切，2表示相交 */

    int operator & (Circle C){

        double d=(c-C.c).length();

        if(dcmp(d)==0){

            if(dcmp(r-C.r)==0) return -1;

            return 0;

        }

        if(dcmp(r+C.r-d)<0) return 0;

        if(dcmp(fabs(r-C.r)-d)>0) return 0;

        double a=(C.c-c).angle();

        double da=acos((r*r+d*d-C.r*C.r)/(2*r*d));

        Point p1=GetPoint(a-da),p2=GetPoint(a+da);

        if(p1==p2) return 1;

        return 2;

    }

};



int n;

int f[N];

Line up,down;

Line lft,rgt;

Circle c[N];



void init()

{

    for(int i=0;i<=n+1;i++) f[i]=i;

    up=Line(Point(0,1000),Point(1000,1000));

    down=Line(Point(0,0),Point(1000,0));

    lft=Line(Point(0,0),Point(0,1000));

    rgt=Line(Point(1000,0),Point(1000,1000));

}

int Find(int x)

{

    if(x!=f[x]) f[x]=Find(f[x]);

    return f[x];

}

void UN(int x,int y)

{

    x=Find(x);

    y=Find(y);

    if(x!=y) f[x]=y;

}

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        init();

        for(int i=1;i<=n;i++) scanf("%lf%lf%lf",&c[i].c.x,&c[i].c.y,&c[i].r);

        for(int i=1;i<=n;i++){

            for(int j=i+1;j<=n;j++){

                if((c[i]&c[j])!=0){

                    UN(i,j);

                }

            }

        }

        //上边界

        for(int i=1;i<=n;i++){

            pair<int,vector<Point> > res=c[i].CircleInterLine(up);

            if(res.first!=0) UN(0,i);

        }

        //下边界

        for(int i=1;i<=n;i++){

            pair<int,vector<Point> > res=c[i].CircleInterLine(down);

            if(res.first!=0) UN(i,n+1);

        }

        if(Find(0)==Find(n+1)){ //出不去

            printf("Bill will be bitten.\n");

            continue;

        }

        //左右边界

        vector<Point> p1,p2;

        p1.push_back(Point(0,1000));

        p2.push_back(Point(1000,1000));

        for(int i=1;i<=n;i++){

            pair<int,vector<Point> > res1=c[i].CircleInterLine(lft);

            pair<int,vector<Point> > res2=c[i].CircleInterLine(rgt);

            if(res1.first!=0){

                while(!res1.second.empty()){

                    if(res1.second.back().y>=0 && res1.second.back().y<=1000 && Find(i)==Find(0))

                        p1.push_back(res1.second.back());

                    res1.second.pop_back();

                }

            }

            if(res2.first!=0){

                while(!res2.second.empty()){

                    if(res2.second.back().y>=0 && res2.second.back().y<=1000 && Find(i)==Find(0))

                        p2.push_back(res2.second.back());

                    res2.second.pop_back();

                }

            }

        }

        int i,j;

        sort(p1.begin(),p1.end());

        sort(p2.begin(),p2.end());

        printf("Bill enters at (0.00, %.2f) and leaves at (1000.00, %.2f).\n",p1[0].y,p2[0].y);

    }

    return 0;

}