Pick-up sticks
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 12861   Accepted: 4847

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.

Source


还是判断线段相交
注意两条线段共线的情况,用点积判断
太诡异从后往前暴力判断能过,用一个栈维护当前没有覆盖的线段也能过
但是最坏复杂度都是N^2啊???
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N=1e5+;
const double eps=1e-;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-; c=getchar();}
while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
return x*f;
}
inline int sgn(double x){
if(abs(x)<eps) return ;
else return x<?-:;
}
struct Vector{
double x,y;
Vector(double a=,double b=):x(a),y(b){}
bool operator <(const Vector &a)const{
return x<a.x||(x==a.x&&y<a.y);
}
void print(){
printf("%lf %lf\n",x,y);
}
};
typedef Vector Point;
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 b){return Vector(a.x*b,a.y*b);}
Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==&&sgn(a.y-b.y)==;} double Cross(Vector a,Vector b){
return a.x*b.y-a.y*b.x;
}
double Dot(Vector a,Vector b){
return a.x*b.x+a.y*b.y;
}
double DisPP(Point a,Point b){
Point t=a-b;
return sqrt(t.x*t.x+t.y*t.y);
}
struct Line{
Point s,t;
Line(){}
Line(Point p,Point v):s(p),t(v){}
}l[N];
bool isLSI(Line l1,Line l2){
Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s;
return sgn(Cross(v,u))!=sgn(Cross(v,w));
}
bool isSSI(Line l1,Line l2){
Vector v1=l1.t-l1.s,v2=l2.t-l2.s;
if(sgn(Cross(v1,v2))==){
int flag=;
Vector u=l2.s-l1.s,w=l2.t-l1.s;
if(sgn(Dot(u,w))<) flag=;
u=l2.s-l1.t,w=l2.t-l1.t;
if(sgn(Dot(u,w))<) flag=;
return flag;
}
else return isLSI(l1,l2)&&isLSI(l2,l1);
} int n;
bool vis[N];
double x,y,x2,y2;
int main(int argc, const char * argv[]) {
while(true){
memset(vis,,sizeof(vis));
n=read(); if(n==) break;
for(int i=;i<=n;i++){
scanf("%lf%lf%lf%lf",&x,&y,&x2,&y2);
l[i]=Line(Point(x,y),Point(x2,y2));
}
for(int i=;i<=n;i++){
for(int j=i+;j<=n;j++) if(isSSI(l[j],l[i])){vis[i]=;break;}
}
printf("Top sticks: ");
int fir=;
for(int i=;i<=n;i++) if(!vis[i]){
if(fir) printf("%d",i),fir=;
else printf(", %d",i);
}
puts(".");
}
return ;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N=1e5+;
const double eps=1e-;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-; c=getchar();}
while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
return x*f;
}
inline int sgn(double x){
if(abs(x)<eps) return ;
else return x<?-:;
}
struct Vector{
double x,y;
Vector(double a=,double b=):x(a),y(b){}
bool operator <(const Vector &a)const{
return x<a.x||(x==a.x&&y<a.y);
}
void print(){
printf("%lf %lf\n",x,y);
}
};
typedef Vector Point;
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 b){return Vector(a.x*b,a.y*b);}
Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==&&sgn(a.y-b.y)==;} double Cross(Vector a,Vector b){
return a.x*b.y-a.y*b.x;
}
double Dot(Vector a,Vector b){
return a.x*b.x+a.y*b.y;
}
double DisPP(Point a,Point b){
Point t=a-b;
return sqrt(t.x*t.x+t.y*t.y);
}
struct Line{
Point s,t;
Line(){}
Line(Point p,Point v):s(p),t(v){}
}l[N];
bool isLSI(Line l1,Line l2){
Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s;
return sgn(Cross(v,u))!=sgn(Cross(v,w));
}
bool isSSI(Line l1,Line l2){
Vector v1=l1.t-l1.s,v2=l2.t-l2.s;
if(sgn(Cross(v1,v2))==){
int flag=;
Vector u=l2.s-l1.s,w=l2.t-l1.s;
if(sgn(Dot(u,w))<) flag=;
u=l2.s-l1.t,w=l2.t-l1.t;
if(sgn(Dot(u,w))<) flag=;
return flag;
}
else return isLSI(l1,l2)&&isLSI(l2,l1);
} int n,st[N],top;
inline void del(int p){
for(int i=p;i<=top;i++) st[i]=st[i+];top--;
}
double x,y,x2,y2;
int main(int argc, const char * argv[]) {
while(true){
top=;
n=read(); if(n==) break;
for(int i=;i<=n;i++){
scanf("%lf%lf%lf%lf",&x,&y,&x2,&y2);
l[i]=Line(Point(x,y),Point(x2,y2));
for(int j=;j<=top;j++) if(isSSI(l[st[j]],l[i])) del(j),j--;
st[++top]=i;
}
printf("Top sticks: %d",st[]);
for(int i=;i<=top;i++) printf(", %d",st[i]);
puts(".");
}
return ;
}

POJ 2653 Pick-up sticks [线段相交 迷之暴力]的更多相关文章

  1. 【POJ 2653】Pick-up sticks 判断线段相交

    一定要注意位运算的优先级!!!我被这个卡了好久 判断线段相交模板题. 叉积,点积,规范相交,非规范相交的简单模板 用了“链表”优化之后还是$O(n^2)$的暴力,可是为什么能过$10^5$的数据? # ...

  2. POJ 2653 Pick-up sticks(线段相交)

    题意:给定n个木棍依次放下,要求最终判断没被覆盖的木棍是哪些. 思路:快速排斥以及跨立实验可以判断线段相交. #include<algorithm> #include<cstdio& ...

  3. POJ 1066 Treasure Hunt (线段相交)

    题意:给你一个100*100的正方形,再给你n条线(墙),保证线段一定在正方形内且端点在正方形边界(外墙),最后给你一个正方形内的点(保证不再墙上) 告诉你墙之间(包括外墙)围成了一些小房间,在小房间 ...

  4. POJ 1410 Intersection --几何,线段相交

    题意: 给一条线段,和一个矩形,问线段是否与矩形相交或在矩形内. 解法: 判断是否在矩形内,如果不在,判断与四条边是否相交即可.这题让我发现自己的线段相交函数有错误的地方,原来我写的线段相交函数就是单 ...

  5. POJ 1269 Intersecting Lines(线段相交,水题)

    id=1269" rel="nofollow">Intersecting Lines 大意:给你两条直线的坐标,推断两条直线是否共线.平行.相交.若相交.求出交点. ...

  6. POJ 1066 Treasure Hunt【线段相交】

    思路:枚举四边墙的门的中点,与终点连成一条线段,判断与其相交的线段的个数.最小的加一即为答案. 我是傻逼,一个数组越界调了两个小时. #include<stdio.h> #include& ...

  7. poj 1556 (Dijkstra + Geometry 线段相交)

    链接:http://poj.org/problem?id=1556 The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissi ...

  8. POJ 3304 Segments[直线与线段相交]

    Segments Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13514   Accepted: 4331 Descrip ...

  9. POJ 1408 Fishnet【枚举+线段相交+叉积求面积】

    题目: http://poj.org/problem?id=1408 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#probl ...

随机推荐

  1. C语言中%d,%p,%u,%lu等都有什么用处

    %d 有符号10进制整数(%ld 长整型,%hd短整型 )%hu 无符号短整形(%u无符号整形,%lu无符号长整形)%i 有符号10进制整数 (%i 和%d 没有区别,%i 是老式写法,都是整型格式) ...

  2. 分布式监控系统--zabbix

    1Zabbix简介 Zabbix 是一个企业级的分布式开源监控方案. 2.监控系统架构 C/S架构 客户端/服务器端,这种架构适合规模较小,处于同一地域的环境 C/P/S 客户端/代理端/服务器端/, ...

  3. [OpenCV学习笔记2][Mat数据类型和操作]

    [Mat数据类型和基本操作] ®.运行环境:Linux(RedHat+OpenCV3.0) 1.Mat的作用: Mat类用于表示一个多维的单通道或者多通道的稠密数组.能够用来保存实数或复数的向量.矩阵 ...

  4. 什么是命名空间?php命名空间的基本应用分享

    什么是命名空间? php中声明的函数名.类名和常量的名称,在同一次运行中是不能重复的,否则会产生一个致命的错误,常见的解决方法是约定一个前缀.例如 ,在项目开发时,用户 User 模块中的控制器和数据 ...

  5. Document类型知识大全

    Document类型 1.文档的子节点  Document类型可以表示HTML页面或者其他基于XML的文档.不过,最常见的应用还是作为HTMLDocument实例的document对象.通过这个文档对 ...

  6. 【自制工具类】Java删除字符串中的元素

    这几天做项目需要把多个item的id存储到一个字符串中,保存进数据库.保存倒是简单,只需要判断之前是否为空,如果空就直接添加,非空则拼接个"," 所以这个字符串的数据结构是这样的 ...

  7. 客户端怎么查看SVN的代码库

    安装SVN客户端,比如TortoiseSVN,然后将代码库checkout到本地,或者通过客户端的版本库浏览器直接连接SVN服务器查看代码库的目录结构. 如果SVN服务器端安装的时候是和Apache集 ...

  8. ip 百度地图 php

    已知一个IP $ipname=api_hits($DT_IP); -------------- //apifunction getAddressComponent($ak, $longitude, $ ...

  9. 【开发技术】storyboard和nib的差别

    在使用Storyboard管理的iOS应用中,它的组成部分为AppDelegate和ViewController这两个类以及MainStoryboard.storyboard文件组成.Storyboa ...

  10. jstl 的判断使用

    JSTL  是JSP的标准标记库 1.必须引入的头部标签 <%@ taglib uri="http://java.sun.com/jstl/core_rt"prefix=&q ...