A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes u and v (u ≠ v) exists either an edge going from u to v, or an edge from v to u. You are give…

第3章 k近邻法   1．近邻法是基本且简单的分类与回归方法.近邻法的基本做法是:对给定的训练实例点和输入实例点,首先确定输入实例点的个最近邻训练实例点,然后利用这个训练实例点的类的多数来预测输入实例点的类. 2．近邻模型对应于基于训练数据集对特征空间的一个划分.近邻法中,当训练集.距离度量.值及分类决策规则确定后,其结果唯一确定. 3．近邻法三要素:距离度量.值的选择和分类决策规则.常用的距离度量是欧氏距离及更一般的pL距离.值小时,近邻模型更复杂:值大时,近邻模型更简单.值的选择反映了对近似…

最近邻法和k-近邻法 下面图片中只有三种豆,有三个豆是未知的种类,如何判定他们的种类? 提供一种思路,即:未知的豆离哪种豆最近就认为未知豆和该豆是同一种类.由此,我们引出最近邻算法的定义:为了判定未知样本的类别,以全部训练样本作为代表点,计算未知样本与所有训练样本的距离,并以最近邻者的类别作为决策未知样本类别的唯一依据.但是,最近邻算法明显是存在缺陷的,比如下面的例子:有一个未知形状(图中绿色的圆点),如何判断它是什么形状? 显然,最近邻算法的缺陷--对噪声数据过于敏感,为了解决这个问题,我们可…

1. All X1.1 基本思路k和c的范围都不大,因此可以考虑迭代找循环节,然后求余数,判定是否相等.这题还是挺简单的.1.2 代码 /* 5690 */ #include <iostream> #include <sstream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <…

import java.util.ArrayList; import java.util.List; /** * Source : https://oj.leetcode.com/problems/permutation-sequence/ * * * The set [1,2,3,-,n] contains a total of n! unique permutations. * * By listing and labeling all of the permutations in orde…

题目大意:维护一个长度为 N 的序列,支持两种操作:区间加,区间查询有多少数是 7 的倍数. 题解:在每个线段树中维护一个权值数组 [0,6],由于个数显然支持区间可加性,因此可用线段树来维护. 代码如下 #include <bits/stdc++.h> using namespace std; const int maxn=1e5+10; inline int read(){ int x=0,f=1;char ch; do{ch=getchar();if(ch=='-')f=-1;}whil…

题意: 平面上给出一个\(N\)个点\(M\)条边的无向图,要用\(C\)种颜色去给每个顶点染色. 如果一种染色方案可以旋转得到另一种染色方案,那么说明这两种染色方案是等价的. 求所有染色方案数 \(mod \: 10^9+7\) 分析: 这种等价类计数的问题可以用Polya定理来解决. 首先这个图形要想能旋转,本身必须中心对称,即旋转以后的顶点和边要和原图完全重合,一一对应. 事实上,旋转的角度只能是\(90^{\circ}\)的整数倍. 因为给出来的点都是整点,求出来的对称中心的坐标也都是有…

Django-Model操作数据库(增删改查.连表结构) 一.数据库操作 1.创建model表        …

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? Linked List Two Pointers     ''' Created on Nov 13, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com> ''' # De…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 思路: 做过,就当复习了. 先用快慢指针判断相交,关键是环开始点的获取. 用上图说明一下,设非环的部分长度为a(包括环的入口点), 环的长度为b(包括环的入口点). 快慢指针相交的位置为绿色的点,距离…

Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree. For example: Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return tru…

http://acm.hdu.edu.cn/showproblem.php?pid=1700 Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1567    Accepted Submission(s): 570 Problem Description There is a cycle with its c…

Description Cycle shifting refers to following operation on the sting. Moving first letter to the end and keeping rest part of the string. For example, apply cycle shifting on ABCD will generate BCDA. Given any two strings, to judge if arbitrary time…

题意:给出n个点m条边的加权有向图,求平均值最小的回路 自己想的是用DFS找环(真是too young),在比较找到各个环的平均权值,可是代码实现不了,觉得又不太对 后来看书= =好巧妙的办法, 使用二分法求解,首先记录下来这m条边的最大权值ub 然后可以猜测一个mid,只需要判断是否存在平均值小于mid的回路 假设存在一个包含k条边的回路,回路上各条边的权值分别为w1,w,2,w3,----,wk 那么 w1+w2+w3+----+wk<k*mid 又因为联想到Bellman_Ford可以解决…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 思考:第一步:设环长为len,快慢指针q.p相遇时,q比p多走了k*len.        第二部:p先走k*len步,p,q一起走每次一步,相遇时p比q多走了一个环距离,此时q就是环开始结点. 通…

题意: 给定一个n个点m条边的带权有向图,求平均权值最小的回路的平均权值? 思路: 首先,图中得有环的存在才有解,其次再解决这个最小平均权值为多少.一般这种就是二分猜平均权值了,因为环在哪也难以找出来,还有可能是一条边属于多个环.对于每个猜到的平均值,如果对应环的存在,那么这个环的每条边的权减去这个平均值之后,用spfa算法就能判断其是否有环的存在即可. 假设环上各边权值为:w1+w2+...+wk. 式子:w1+w2+...+wk<k*even   相当于   (w1-even)+(w2-ev…

这题的俩种方法都是看别人的代码,方法可以学习学习,要多看看.. 几何题用到向量.. Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1294    Accepted Submission(s): 455 Problem Description There is a cycle with its cente…

LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Linked List Cycle II Given a linked list, return the node w…

证明单链表有环路: 本文所用的算法 能够 形象的比喻就是在操场其中跑步.速度快的会把速度慢的扣圈  能够证明,p2追赶上p1的时候.p1一定还没有走完一遍环路,p2也不会跨越p1多圈才追上  我们能够从p2和p1的位置差距来证明.p2一定会赶上p1可是不会跳过p1的  由于p2每次走2步.而p1走一步.所以他们之间的差距是一步一步的缩小,4,3,2,1,0  到0的时候就重合了. 找到环路的起点: 既然可以推断出是否是有环路,那改怎样找到这个环路的入口呢?  解法例如以下: 当p2依照每次2步,…

Problem Description There is a cycle with its center on the origin. Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other you may assume that the radius of the cycle…

option=com_onlinejudge&Itemid=8&page=show_problem&problem=4336" style="">题目链接:uva 1561 - Cycle Game 题目大意:给出一个环,每次从起点開始,能够选择一个权值非0的边移动,移动后减掉权值至少1点. 不能移动的为失败. 解题思路: 1:有0的情况,假设有方向离权值为0的边的步数为奇数,则为必胜.否则必败. 2:无0的情况,奇数边必胜: 3:有1的情况.同…

Problem G: Going in Cycle!! Input: standard input Output: standard output You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the gra…

1.Linked List Cycle 题目链接 题目要求: Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 刚看到这道题,很容易写出下边的程序: bool hasCycle(ListNode *head) { ListNode *a = head, *b = head; while(a) { b = a->next; wh…

1122. Hamiltonian Cycle (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle"…

Python交互K线工具 K线核心功能+指标切换 aiqtt团队量化研究,用vn.py回测和研究策略.基于vnpy开源代码,刚开始接触pyqt,开发界面还是很痛苦,找了很多案例参考,但并不能完全满足我们自己对于检查自己的交易逻辑的需求,只能参考网上的案例自己开发 代码较多,大家可以直接到GitHub下载开源源码查看 欢迎加入QQ交流群: 538665416(免费提供,期货,期权数据) 团队界面需求: 界面加载k线, 鼠标滚轮缩放,键盘缩放跳转 十字光标 显示K线详细信息 缩放自适应Y轴坐标 回测…

The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle". In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle. Input Specif…

题目链接:http://codeforces.com/gym/101981/problem/K Your friend has made a computer video game called “Kangaroo Puzzle” and wants you to give it a try for him. As the name of this game indicates, there are some (at least 2) kangaroos stranded in a puzzle…

[题目描述] 给定一个 n 个点 m 条边的加权有向图,求平均权值最小的回路. [输入格式] 输入第一行为数据组数 T .每组数据第一行为图的点数 n 和边数 m (n ≤ 50).以下 m 行每行3个整数 u, v, w, 表示有一条从 u 到 v 的有向边,权值为 w.输入没有自环. [输出格式] 对于每组数据,输出平均最小值,并保留2位小数.如果误解,输出 "No cycle found.". 这道题吧,我觉得使用二分法求解不错.首先才一个值 mid,只需要判断是否存在平均值小于…

The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle". In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle. Input Specif…

1122 Hamiltonian Cycle (25 分) The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle". In this problem, you are supposed to tell if a given cycle is a H…