id=2002">Squares 很好的一道二分,事实上本来我是没有思路的,看了基神的题解之后才似乎明确了点. 题意:给出最多有1000个点,问这些点能够组成多少个正方形 分析:先想想怎么推断正方形吧.假如我如今有四个点A,B,C,D,由于边不一定是平行于x轴的.可能是倾斜的,怎样推断四个点组成的四边形是正方形呢?以A点为中心,AB为轴,绕A点顺时针或者逆时针转动90度,就能够得到一个点D' , 相同的方法.以B 点为中心,BA为轴.绕B点顺时针或者逆时针转动90度.就能够得到一个点C'…

http://poj.org/problem?id=2002 只能说hash比二分快很多.随便一个hash函数都可以完爆二分. 判断是否存在正方形思路如下: 1.枚举任意两个点,作为正方形的一条边,那么,整个正方形就确定了,有两个方向. 因为, 设枚举的坐标为(x1, y1) & (x2, y2),所求的坐标是和x1,y1这个点相连,那么有方程如下. 1.垂直,向量积是0 2.边长相等,然后距离公式化简. 即可解出剩下的两个点. 然后要注意两个点要在正方形的同一侧,不然变了平行四边形了. 唤醒了…

Squares Time Limit: 3500MS   Memory Limit: 65536K Total Submissions: 16631   Accepted: 6328 Description A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating abou…

题目链接: http://poj.org/problem?id=2002 #include <stdio.h> #include <string.h> ; struct Hash_table { int x, y; struct Hash_table *next; }*Hash[prime]; void Hash_insert(int x, int y) { int key = (x + y) % prime; if(Hash[key] == NULL) { Hash[key] =…

Squares Time Limit: 3500MS   Memory Limit: 65536K Total Submissions: 15137   Accepted: 5749 Description A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating abou…

经典好题. 题意是要我们找出所有的正方形.1000点,只有枚举咯. 如图,如果我们知道了正方形A,B的坐标,便可以推测出C,D两点的坐标.反之,遍历所有点作为A,B点,看C,D点是否存在.存在的话正方形数+1. 假设A点坐标为(x1,y1),B点坐标为(x2,y2),则根据三角形全等,易知 C点坐标:( x1+(y2-y1),y1-(x2-x1) ) D点坐标:( x2+(y2-y1),y2-(x2-x1) ) 当然,如果我们遍历任意两个点,一个正方形将会被计数四次(四条边).我们可以只判断斜率…

题目 http://poj.org/problem?id=2002 题意 已知平面内有1000个点,所有点的坐标量级小于20000,求这些点能组成多少个不同的正方形. 思路 如图,将坐标按照升序排列后,首先枚举p1,p2, 并判断p2是否在p1正下方或者左上角(因为每个正方形只有一条最右边或者是右下的边),按照下图计算p3,p4,判断p3,p4是否存在即可. 感想 排序时要注意和左上角这个信息相符,刚写完时用的是左下角,与升序排序不符合,会遗失部分正方形. 代码 #include <cstdio…

Squares Time Limit: 3500MS   Memory Limit: 65536K Total Submissions: 18493   Accepted: 7124 Description A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating abou…

二分.... Squares Time Limit: 3500MS Memory Limit: 65536K Total Submissions: 14530 Accepted: 5488 Description A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating a…

body{ background-color: #000; margin:; padding:; } main{ perspective: 800px; } .cube{ transform-style: preserve-3d; position: relative; margin: 200px auto 0px; width: 400px; height: 400px; animation: spin 8s linear infinite; animation-play-state: pau…

/** * 图片绕边旋转的小动画 */ function initDemo10() { var canvas = document.getElementById("demo10"); if (!canvas) { return; } var context = canvas.getContext("2d"); var img = new Image(); img.src ="images/timg3.jpg"; img.onload = func…

枚举两个点作为一条边,求出正方形的另外两个点,利用hash查找另外两个点. #include<stdio.h> #include<string.h> #include<stdlib.h> struct node{ int x,y; } point[]; struct hash{ int pos; hash* next; } hashtable[]; int find(int x,int y){ hash* t; ; t=&hashtable[tmp]; ){ ;…

题目链接:http://poj.org/problem?id=2002 题意:给定n个点,问有多少种方法可以组成正方形. 思路:我们可以根据两个点求出对应正方形[有2个一个在两点左边,一个在两点右边]另外两个点的左边.例如 已知:(x1,y1) (x2,y2)则x3=x1+(y1-y2) y3= y1-(x1-x2) x4=x2+(y1-y2) y4= y2-(x1-x2)或x3=x1-(y1-y2) y3= y1+(x1-x2) x4=x2-(y1-y2) y4= y2+(x1-x2) 枚举两…

Squares Time Limit: 3500MS   Memory Limit: 65536K Total Submissions: 15489   Accepted: 5864 Description A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating abou…

题目大意: 给定1000个点,寻找有多少组四点对能组成正方形 这里的题目跟上一道做的找平行四边形类似但想法却又不相同的方法 这里找任意2个点形成的一条边,那么可以根据这两个点,找到能和他们组成正方形剩下的两个点的位置,根据hash表去搜索,如果这两个位置存在自己需要的点,说明这种方案可行 添加查找均交给hash表,这样可以实现O(n*n)的复杂度 最后因为每个正方形因为有4条边被访问了4次,所以总的答案要除以4 #include <cstdio> #include <cstring>…

题目链接: https://vjudge.net/problem/POJ-2002 题目大意: 有一堆平面散点集,任取四个点,求能组成正方形的不同组合方式有多少. 相同的四个点,不同顺序构成的正方形视为同一正方形. 解题思路: 直接四个点四个点地枚举肯定超时的,不可取. 普遍的做法是:先枚举两个点(这两个点是正方形的一条边),通过数学公式得到另外2个点,使得这四个点能够成正方形.然后检查散点集中是否存在计算出来的那两个点,若存在,说明有一个正方形. 但这种做法会使同一个正方形按照不同的顺序被枚举…

[题目链接] http://poj.org/problem?id=2758 [题目大意] 给出一个字符串,支持两个操作,在任意位置插入一个字符串,或者查询两个位置往后的最长公共前缀,注意查询的时候是原串下标,插入的时候则是最近更新串的下标. [题解] 因为插入操作只有两百次,所以考虑hash重构来处理匹配问题,碰到插入就重构插入点往后的哈希表,否则二分两个位置往后的匹配长度,查hash表判断是否可行. [代码] #include <cstdio> #include <algorithm&…

题目链接:http://poj.org/problem?id=2187 Time Limit: 3000MS Memory Limit: 65536K Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of…

最近用到了QGraphicsItem,可以通过QGraphicsItemAnimation使其产生动画效果. QGraphicsItemAnimation自带了setPosAt().setRotationAt().setScaleAt()等方法可以用来移动.旋转.放缩QGraphicsItem,但其默认的OriginPoint是这个Item的左上角,虽然QGraphicsItem自带了setTransformOriginPoint()方法,但是设置以后没有效果,还是绕左上角放缩旋转,只好采取其他…

Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15900   Accepted: 5494 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the…

链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/E Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24254   Accepted: 7403 Description Bessie, Farmer John's prize cow, h…

n个点 求其中有几个正方形 n<1000 暴力4个点就不行了 大概2个点还可以 根基(x*x+y*y)%素数 hash 一下 告诉你2个点求 另外2个点 画个图推一下 重复要/2; #include<stdio.h> #include<algorithm> #include<vector> using namespace std; #define mod 10007 struct point { int x,y; }x[]; vector<point>…

题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=17267 [思路] 凸包+旋转卡壳 求出凸包,用旋转卡壳算出凸包的直径即可. [代码] #include<cstdio> #include<vector> #include<iostream> #include<algorithm> using namespace std; struct Pt { int x,y; Pt(,):…

1. 之前有放过一个初始版本,但是因为在旋转的时候,有比较大的瑕疵,造成每个点运动到端点后,出现类似撞击的感觉. 2. 所以本文对旋转作了些调整,运用类似水平方向的圆周运动 a. HTML代码,定义canvas标签 <!DOCTYPE html> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content…

写下来发现,程序在细节上处理的很差,比如旋转的时候,在终点处有明显的撞墙感觉,以及小部分粒子存在精度差异,导致撞击后不与整体平衡. 注释全在代码中了,就不多说了,另外感觉写的旋转的规则有点怪,后续再调整吧. <!DOCTYPE html> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content=&qu…

Description Dao was a simple two-player board game designed by Jeff Pickering and Ben van Buskirk at . A variation of it, called S-Dao, * square with cells. There are black stones and white stones placed on the game board randomly in the beginning. T…

Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10455   Accepted: 3093   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…

Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 48646   Accepted: 12703 Description You may have heard that no two snowflakes are alike. Your task is to write a program to determine whether this is really true.…

题目大意:原题链接 给定平面上的N个点,求出这些点一共可以构成多少个正方形. 解题思路: 若正方形为ABCD,A坐标为(x1, y1),B坐标为(x2, y2),则很容易可以推出C和D的坐标.对于特定的A和B坐标,C和D可以在线段AB的上面或者下面,即有两种情况.           根据构造三角形全等可以得知(很简单,注意下细节,不要把坐标弄混就行) CD在AB上面x3=x2+(y1-y2),y3=y2+(x2-x1);    x4=x1+(y1-y2),y4=y1+(x2-x1); CD在A…

Description Fractals are really cool mathematical objects. They have a lot of interesting properties, often including: 1. fine structure at arbitrarily small scales; 2. self-similarity, i.e., magnified it looks like a copy of itself; 3. a simple, rec…