首页 > 代码库 > POJ 2653 Pick-up sticks --队列,几何
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-8using namespace std;#define N 100017struct 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 segmentstruct node{ Point P[2];}p[N];vector<int> G;//data endsint 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;}
POJ 2653 Pick-up sticks --队列,几何
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。