Pick-up sticks

Pick-up sticks

http://poj.org/problem?id=2653

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15265		Accepted: 5752

Description

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.

Input

Input consists of a number of cases. The data for each case start with 1 <= n <= 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.

Output

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown. 

The picture to the right below illustrates the first case from input.

Sample Input

5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0

Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

Hint

Huge input,scanf is recommended.

 

这题判断是否相交就行

把bin巨的模板打了一遍。。。

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const double eps=1e-8;
const double INF=1e20;
const double PI=acos(-1.0);
const int maxp=1010;
int sgn(double x){
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    else return 1;
}
inline double sqr(double x){return x*x;}
struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x=_x;
        y=_y;
    }
    void input(){
        scanf("%lf %lf",&x,&y);
    }
    void output(){
        printf("%.2f %.2f\n",x,y);
    }
    bool operator == (const Point &b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y)== 0;
    }
    bool operator < (const Point &b)const{
        return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator - (const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^ (const Point &b)const{
        return x*b.y-y*b.x;
    }
    //点积
    double operator * (const Point &b)const{
        return x*b.x+y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);
    }
    //返回长度的平方
    double len2(){
        return x*x+y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator + (const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k,y/k);
    }

    //计算pa和pb的夹角
    //就是求这个点看a,b所成的夹角
    ///LightOJ1202
    double rad(Point a,Point b){
        Point p=*this;
        return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
    }
    //化为长度为r的向量
    Point trunc(double r){
        double l=len();
        if(!sgn(l)) return *this;
        r/=l;
        return Point(x*r,y*r);
    }
    //逆时针转90度
    Point rotleft(){
        return Point(-y,x);
    }
    //顺时针转90度
    Point rotright(){
        return Point(y,-x);
    }
    //绕着p点逆时针旋转angle
    Point rotate(Point p,double angle){
        Point v=(*this) -p;
        double c=cos(angle),s=sin(angle);
        return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
    }
};

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s=_s;
        e=_e;
    }
    bool operator==(Line v){
        return (s==v.s)&&(e==v.e);
    }
    //根据一个点和倾斜角angle确定直线，0<=angle<pi
    Line(Point p,double angle){
        s=p;
        if(sgn(angle-PI/2)==0){
            e=(s+Point(0,1));
        }
        else{
            e=(s+Point(1,tan(angle)));
        }
    }
    //ax+by+c=0;
    Line(double a,double b,double c){
        if(sgn(a)==0){
            s=Point(0,-c/b);
            e=Point(1,-c/b);
        }
        else if(sgn(b)==0){
            s=Point(-c/a,0);
            e=Point(-c/a,1);
        }
        else{
            s=Point(0,-c/b);
            e=Point(1,(-c-a)/b);
        }
    }
    void input(){
        s.input();
        e.input();
    }
    void adjust(){
        if(e<s) swap(s,e);
    }
    //求线段长度
    double length(){
        return s.distance(e);
    }
    //返回直线倾斜角 0<=angle<pi
    double angle(){
        double k=atan2(e.y-s.y,e.x-s.x);
        if(sgn(k)<0) k+=PI;
        if(sgn(k-PI)==0) k-=PI;
        return k;
    }
    //点和直线的关系
    //1 在左侧
    //2 在右侧
    //3 在直线上
    int relation(Point p){
        int c=sgn((p-s)^(e-s));
        if(c<0) return 1;
        else if(c>0) return 2;
        else return 3;
    }
    //点在线段上的判断
    bool pointonseg(Point p){
        return sgn((p-s)^(e-s))==0&&sgn((p-s)*(p-e))<=0;
    }
    //两向量平行(对应直线平行或重合)
    bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s))==0;
    }
    //两线段相交判断
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(Line v){
        int d1=sgn((e-s)^(v.s-s));
        int d2=sgn((e-s)^(v.e-s));
        int d3=sgn((v.e-v.s)^(s-v.s));
        int d4=sgn((v.e-v.s)^(e-v.s));
        if((d1^d2)==-2&&(d3^d4)==-2) return 2;
        return (d1==0&&sgn((v.s-s)*(v.s-e))<=0||
                d2==0&&sgn((v.e-s)*(v.e-e))<=0||
                d3==0&&sgn((s-v.s)*(s-v.e))<=0||
                d4==0&&sgn((e-v.s)*(e-v.e))<=0);
    }
    //直线和线段相交判断
    //-*this line -v seg
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int linecrossseg(Line v){
        int d1=sgn((e-s)^(v.s-s));
        int d2=sgn((e-s)^(v.e-s));
        if((d1^d2)==-2) return 2;
        return (d1==0||d2==0);
    }
    //两直线关系
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(Line v){
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    //求两直线的交点
    //要保证两直线不平行或重合
    Point crosspoint(Line v){
        double a1=(v.e-v.s)^(s-v.s);
        double a2=(v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
    //点到直线的距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();
    }
    //点到线段的距离
    double dispointtoseg(Point p){
        if(sgn((p-s)*(e-s))<0||sgn((p-e)*(s-e))<0)
            return min(p.distance(s),p.distance(e));
        return dispointtoline(p);
    }
    //返回线段到线段的距离
    //前提是两线段不相交，相交距离就是0了
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
    }
    //返回点P在直线上的投影
    Point lineprog(Point p){
        return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
    }
    //返回点P关于直线的对称点
    Point symmetrypoint(Point p){
        Point q=lineprog(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
};

Line L[100005];
int book[100005];
int n;

bool Check(Line a,Line b){
    if(((a.s-a.e)^(b.s-a.e))*((a.s-a.e)^(b.e-a.e))>0) return false;
    if(((b.s-b.e)^(a.s-b.e))*((b.s-b.e)^(a.e-b.e))>0) return false;
    return true;
}


int main(){
    while(~scanf("%d",&n)){
        if(!n) break;
        memset(book,0,sizeof(book));
        for(int i=1;i<=n;i++){
            scanf("%lf %lf %lf %lf",&L[i].s.x,&L[i].s.y,&L[i].e.x,&L[i].e.y);
        }
        int flag;
        for(int i=1;i<=n;i++){
            flag=0;
            for(int j=i+1;j<=n;j++){
                if(Check(L[i],L[j])){
                    flag=1;
                    break;
                }
            }
            if(flag){
                book[i]=1;
            }
        }
        flag=0;
        printf("Top sticks: ");
        for(int i=1;i<=n;i++){
            if(!book[i]){
                if(!flag) flag=1;
                else printf(", ");
                printf("%d",i);

            }
        }
        printf(".\n");
    }
    return 0;
}