POJ 2653 Pick-up sticks --队列，几何

题意： 按顺序扔木棒，求出最上层的木棒是哪些。

解法： 由于最上层的木棒不超过1000个，所以用一个队列存储最上层的木棒，每次扔出一个木棒后，都与队列中的木棒一一判断，看此木棒是否在某一最上层的木棒的上面，即判线段是否相交（两次跨立实验），如果相交，则将那个被压的木棒抛出队列，最后再加入扔的这个木棒到队列中。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

#define eps 1e-8

using namespace std;

#define N 100017



struct Point{

    double x,y;

    Point(double x=0, double y=0):x(x),y(y) {}

    void input() { scanf("%lf%lf",&x,&y); }

};

typedef Point Vector;

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); }

    void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }

};

struct Line{

    Point p;

    Vector v;

    double ang;

    Line(){}

    Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); }

    Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); }

    bool operator < (const Line &L)const { return ang < L.ang; }

};

int dcmp(double x) {

    if(x < -eps) return -1;

    if(x > eps) return 1;

    return 0;

}

template <class T> T sqr(T x) { return x * x;}

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); }

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 a.x >= b.x && a.y >= b.y; }

bool operator <= (const Point& a, const Point& b) { return 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(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }

double Length(Vector A) { return sqrt(Dot(A, A)); }

double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

Vector VectorUnit(Vector x){ return x / Length(x);}

Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}

double angle(Vector v) { return atan2(v.y, v.x); }



bool OnSegment(Point P, Point A, Point B) {

    return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;

}

double DistanceToSeg(Point P, Point A, Point B)

{

    if(A == B) return Length(P-A);

    Vector v1 = B-A, v2 = P-A, v3 = P-B;

    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);

    if(dcmp(Dot(v1, v3)) > 0) return Length(v3);

    return fabs(Cross(v1, v2)) / Length(v1);

}

double DistanceToLine(Point P, Point A, Point B){

    Vector v1 = B-A, v2 = P-A;

    return fabs(Cross(v1,v2)) / Length(v1);

}

Point GetLineIntersection(Line A, Line B){

    Vector u = A.p - B.p;

    double t = Cross(B.v, u) / Cross(A.v, B.v);

    return A.p + A.v*t;

}

bool SegmentIntersection(Point A,Point B,Point C,Point D) {

    if(dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 && dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0) return true;

    return false;

}

//data segment

struct node{

    Point P[2];

}p[N];

vector<int> G;

//data ends



int main()

{

    int n,i,j;

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

    {

        queue<int> q;

        G.clear();

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

            p[i].P[0].input(), p[i].P[1].input();

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

        {

            Point A = p[i].P[0], B = p[i].P[1];

            int sz = q.size();

            while(sz--)

            {

                int now = q.front();

                q.pop();

                if(!SegmentIntersection(A,B,p[now].P[0],p[now].P[1]))

                    q.push(now);

            }

            q.push(i);

        }

        while(!q.empty())

            G.push_back(q.front()), q.pop();

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

        printf("Top sticks:");

        for(i=0;i<G.size()-1;i++)

            printf(" %d,",G[i]);

        printf(" %d.\n",G[i]);

    }

    return 0;

}

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