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.

 
题目大意:在地上抛n条棍子,问有哪几条棍子没有被其他棍子压着。
思路:暴力搜,这题暂时还没有发现什么有效的解法……
有人说You may assume that there are no more than 1000 top sticks.所以换个姿势暴力就不会TLE了……
明明最坏还是得O(n^2)嘛……只要你给我代码还是能卡……比如我的代码……前n-1个线段都不相交,最后一条跨立前n-1个线段,然后每次都要扫到最后,就卡掉了……
除非上面那句话说的是任何时刻而不是最后时刻,嘛没有说清楚这就是出题人的问题了……(在线计算,拿个1000的队列记住那些还没被覆盖,复杂度O(n * 1000))
反正我只是为做模板而已……随便啦这种小事……
 
代码(516MS):
 #include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std; const double EPS = 1e-; inline int sgn(double x) {
return (x > EPS) - (x < -EPS);
} struct Point {
double x, y;
Point() {}
Point(double x, double y): x(x), y(y) {}
void read() {
scanf("%lf%lf", &x, &y);
}
bool operator < (const Point &rhs) const {
if(y != rhs.y) return y < rhs.y;
return x < rhs.x;
}
Point operator + (const Point &rhs) const {
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point &rhs) const {
return Point(x - rhs.x, y - rhs.y);
}
Point operator * (const int &b) const {
return Point(x * b, y * b);
}
Point operator / (const int &b) const {
return Point(x / b, y / b);
}
double length() const {
return sqrt(x * x + y * y);
}
Point unit() const {
return *this / length();
}
};
typedef Point Vector; double dist(const Point &a, const Point &b) {
return (a - b).length();
} double across(const Point &a, const Point &b) {
return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn left
double cross(const Point &sp, const Point &ed, const Point &op) {
return sgn(across(sp - op, ed - op));
} struct Seg {
Point st, ed;
Seg() {}
Seg(Point st, Point ed): st(st), ed(ed) {}
void read() {
st.read(); ed.read();
}
}; bool isIntersected(Point s1, Point e1, Point s2, Point e2) {
return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
(max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
(max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
(max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
(cross(s2, e1, s1) * cross(e1, e2, s1) >= ) &&
(cross(s1, e2, s2) * cross(e2, e1, s2) >= );
} bool isIntersected(Seg a, Seg b) {
return isIntersected(a.st, a.ed, b.st, b.ed);
} /*******************************************************************************************/ const int MAXN = ; Seg s[MAXN];
bool isAns[MAXN];
Point p;
int n; int main() {
while(scanf("%d", &n) != EOF && n) {
memset(isAns, true, sizeof(isAns));
for(int i = ; i < n; ++i) s[i].read();
for(int i = ; i < n; ++i)
for(int j = i + ; j < n; ++j) {
if(isIntersected(s[i], s[j])) {
isAns[i] = false;
break;
}
}
bool flag = false;
printf("Top sticks:");
for(int i = ; i < n; ++i) {
if(!isAns[i]) continue;
if(flag) printf(",");
else flag = true;
printf(" %d", i + );
}
puts(".");
}
}

POJ 2653 Pick-up sticks(线段判交)的更多相关文章

  1. (线段判交的一些注意。。。)nyoj 1016-德莱联盟

    1016-德莱联盟 内存限制:64MB 时间限制:1000ms 特判: No通过数:9 提交数:9 难度:1 题目描述: 欢迎来到德莱联盟.... 德莱文... 德莱文在逃跑,卡兹克在追.... 我们 ...

  2. (叉积,线段判交)HDU1086 You can Solve a Geometry Problem too

    You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/3 ...

  3. 【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? # ...

  4. POJ 1556 The Doors 线段判交+Dijkstra

    The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6734   Accepted: 2670 Descrip ...

  5. (计算几何 线段判交) 51nod1264 线段相交

    1264 线段相交 给出平面上两条线段的两个端点,判断这两条线段是否相交(有一个公共点或有部分重合认为相交). 如果相交,输出"Yes",否则输出"No".   ...

  6. 线段相交 POJ 2653

    // 线段相交 POJ 2653 // 思路:数据比较水,据说n^2也可以过 // 我是每次枚举线段,和最上面的线段比较 // O(n*m) // #include <bits/stdc++.h ...

  7. poj 2653 线段与线段相交

    Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 11884   Accepted: 4499 D ...

  8. POJ 2653

    题目大意:一个小孩不断往地上扔棍子,共n根,求结束后共有多少根不与去他相交. 解法思路:典型的判断线段相交问题,利用快速排斥+双跨立判断相交,最后输出没相交的. (图片来源:http://www.2c ...

  9. The 2015 China Collegiate Programming Contest D.Pick The Sticks hdu 5543

    Pick The Sticks Time Limit: 15000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others ...

随机推荐

  1. GPUImage源码解读之GPUImageContext

    GPUImageContext类,提供OpenGL ES基本上下文,GPUImage相关处理线程,GLProgram缓存.帧缓存.由于是上下文对象,因此该模块提供的更多是存取.设置相关的方法. 属性列 ...

  2. 浅谈async函数await用法

    今天状态不太好,睡久了懵一天. 以前只是了解过async函数,并还没有很熟练的运用过,所以先开个坑吧,以后再结合实际来更新下,可能说的有些问题希望大家指出. async和await相信大家应该不陌生, ...

  3. Javascript Code Style Guide

    本指南采用的Airbnb发布的基于ES5的JavaScript Code Style. ES5 英文版:https://github.com/airbnb/javascript/tree/es5-de ...

  4. SpringBoot使用maven插件打包時報:[ERROR] [Help 1] http://cwiki.apache.org/confluence/display/MAVEN/PluginExecutionException的處理方案

    SpringBoot使用maven插件打包時報:[ERROR] [Help 1] http://cwiki.apache.org/confluence/display/MAVEN/PluginExec ...

  5. Shell中的${}、##和%%使用范例

    假设定义了一个变量为,代码如下: file=/dir1/dir2/dir3/my.file.txt 可以用${ }分别替换得到不同的值: ${file#*/}: 删掉第一个 / 及其左边的字符串:di ...

  6. Hibernate 事务不回滚

    问题:               这几天在做开发时,发现事务不回滚了,Service是用AOP加的事务,数据库是MySql, 表全部是InnoDB:   方法回滚是采用spring的手动回滚:   ...

  7. 使用Screen管理远程会话

    ​ 在本地开发时,经常需要使用远程连接到Linux服务器,一开始我自己都是有几个远程就开几个SSH窗口,这种方法很原始很直接,但每次都受够了密码输入,即使用了SSH免密码登录,也会觉得每次输入SSH的 ...

  8. 我的Tmux学习笔记

    0. 修改指令前缀 // ~/.tmux.conf ubind C-b set -g prefix C-a 1. 新建会话 tmux tmux new -s session-name // 可以设置会 ...

  9. day 20 约束 异常处理 MD5

    1.类的约束(重点): 写一个父类.  父类中的某个方法要抛出一个异常  NotImplementError # 项目经理 class Base:     # 对子类进行了约束. 必须重写该方法    ...

  10. FPGA软硬协同设计学习笔记及基础知识(一)

    一.FPGA软件硬件协同定义: 软件:一般以软件语言来描述,类似ARM处理器的嵌入式设计.硬件定义如FPGA,里面资源有限但可重配置性有许多优点,新的有动态可充配置技术. Xilinx开发了部分动态可 ...