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? #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 100005 using namespace std; struct Point { double x…

POJ2653 判断线段相交的方法 先判断直线是否相交 再判断点是否在线段上 复杂度是常数的 题目保证最后答案小于1000 故从后往前尝试用后面的线段 "压"前面的线段 排除不可能的答案 就可以轻松AC了. #include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> #include<algorit…

POJ 2826 An Easy Problem?! -- 思路来自kuangbin博客 下面三种情况比较特殊,特别是第三种 G++怎么交都是WA,同样的代码C++A了 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const double eps = 1e-8;…

Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 10330   Accepted: 3833 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 fin…

题目链接:http://poj.org/problem?id=1066 Time Limit: 1000MS Memory Limit: 10000K Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-art technology they are…

黑书上的一道例题:如果走最短路则会碰到点,除非中间没有障碍. 这样把能一步走到的点两两连边,然后跑SPFA即可. #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 100003 using namespace std; struct Point { double x, y; Point(double _x = 0, double _y = 0) :…

Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7699   Accepted: 2843 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…

Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7857   Accepted: 3247 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-ar…

POJ_2653_Pick-up sticks_判断线段相交 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 o…

传统解法 题目来自 leetcode 335. Self Crossing. 题意非常简单,有一个点,一开始位于 (0, 0) 位置,然后有规律地往上,左,下,右方向移动一定的距离,判断是否会相交(self crossing). 一个很容易想到的方案就是求出所有线段,然后用 O(n^2) 的时间复杂度两两判断线段是否相交,而线段相交的判断,可以列个二元一次方程求解(交点).这个解法非常容易想到,但是实际操作起来比较复杂,接下去介绍利用向量的解法. 向量解法 简单回顾下向量,具体的自行谷歌.向量就…