1.数的划分 点击查看搜索 #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; int n,m,a[100]; void dfs(int x,int y,int z)//shu zong wei { if(y>n)return; if(y==n) { for(int i=1;i<z;i++)//不取等 { prin…

http://acm.timus.ru/problem.aspx?space=1&num=1923 -- timus This is s problem about thd dfs and searching like the leetcode islands problems Creating three 2-D arrays: board, land(copy), visit. board: the input, land:(0: undecided, 1: B country, 2: M…

1. Trees Tree is a recursive structure. 1.1 math nodes https://class.coursera.org/principlescomputing-001/wiki/view? page=trees 1.2 CODE无parent域的树 http://www.codeskulptor.org/#poc_tree.py class Tree: """ Recursive definition for trees plus…

Problem: Given a 2D board and a list of words from the dictionary, find all words in the board. Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. Th…

Given an Android 3x3 key ≤ m ≤ n ≤ , count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys. Rules for a valid pattern: Each pattern must connect at least m keys and at most n keys.…

PE刷题记录 PE60 / 20%dif 这道题比较坑爹. 所有可以相连的素数可以构成一张图,建出这张图,在其中找它的大小为5的团.注意上界的估算,大概在1W以内.1W内有1229个素数,处理出这些素数的关系,然后dfs这张图找出大小为5的团. /*** * @name Prime pair sets * @author zball * @algorithm Sieve for primes and dfs search for finding a set of primes */ #inclu…

Given an Android 3x3 key lock screen and two integers m and n, where 1 ≤ m ≤ n ≤ 9, count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys. Rules for a valid pattern: Each pattern m…

Trie build and search class TrieNode { public: TrieNode * next[]; bool is_word; TrieNode(bool b = false) { memset(next,,sizeof(next)); is_word = b; } }; class Trie { TrieNode* root; public: /** Initialize your data structure here. */ Trie() { root =…

Problem statement: Given n processes, each process has a unique PID (process id) and its PPID (parent process id). Each process only has one parent process, but may have one or more children processes. This is just like a tree structure. Only one pro…

Given a binary tree where every node has a unique value, and a target key k, find the value of the nearest leaf node to target k in the tree. Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of th…

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar. For example, words1 = ["great", "acting", "skills"] and words2 = [&…

Given an Android 3x3 key lock screen and two integers m and n, where 1 ≤ m ≤ n ≤ 9, count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys. Rules for a valid pattern: Each pattern m…

本题解同步于luogu emmm切了近年省选题来写题解啦qwq 该题较其他省选题较水吧(否则我再怎么做的出来 思路是图论做法,做法上楼上大佬已经讲的很清楚了,我来谈谈代码实现上的一些细节 \[\text{设节点1...2n,i}\in\text{1-n表示i行,i}\in\text{(n+1)-2n时表示i-n列}\] \[\text{当我们读到一颗绿宝石(x,y,k)时,就从x向y+n连一条权值为k的边}\] \[\text{当我们连完边后会发现给一行/一列增加a就相当于把与这个点相连的所有边…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

Validate Binary Search Tree Total Accepted: 23828 Total Submissions: 91943My Submissions Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with…

There are generally two methods to write DFS algorithm, one is using recursion, another one is using stack. (reference from Wiki Pedia) Pseudocode for both methods: A recursive implementation of DFS: procedure DFS(G,v): label v as discovered for all…

最后一例,搞得快.三天之内走了一次.. 下一步,面象对像的javascript编程. function Dictionary(){ var items = {}; this.has = function (key) { return key in items; }; this.set = function(key, value){ items[key] = value; }; this.remove = function(key){ if (this.has(key)){ delete item…

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key.    The right subtree of a node contains only nodes with k…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. For example, Given n = 3, your program should return all 5 unique BST's shown below. 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3 思路:结果要保留树的所有节点,所以每次都…

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?   法I:BST的中序遍历结果是递增序列.把这个递增序列…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 思路:使用二分法,将list的中间节点作为根节点,然后分别处理list左半边及右半边,以此递归. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right…

DFS主要在于参数的改变; 样例输入: n=4                //给定n个数字 a={1,2,4,7}    //输入n个数据 k=15              //目标数字 样例输出: No 题意: 给定的数字在不重复使用的前提下能否达到目标,能输出Yes,否输出No #include<algorithm> #include<iostream> using namespace std; int n,k,a[10000]; bool dfs(int i,int s…

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. For example,Given n = 3, your program should return all 5 unique BST's shown below. 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3 confused what "{1,#,2…

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. Hide Tags Depth-first Search Linked List       这题是将链表变成二叉树,比较麻烦的遍历过程,因为链表的限制,所以深度搜索的顺序恰巧是链表的顺序,通过设置好递归函数的参数,可以在深度搜索时候便可以遍历了.   TreeNode * he…

题意 略 分析 1.首先要了解到BST的中序遍历是递增序列 2.我们用一个临时节点tmp储存p的中序遍历的下一个节点,如果p->right不存在,那么tmp就是从root到p的路径中大于p->val的最小数,否则就遍历p的右子树,找到最左边的节点即可 代码 class Solution { public: /* * @param root: The root of the BST. * @param p: You need find the successor node of p. * @re…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys…

深度优先搜索是一种枚举所有完整路径以遍历所有情况的搜索方法.(不撞南墙不回头) DFS一般用递归来实现,其伪代码思路过程一般如下: void DFS(必要的参数){    if (符和遍历到一条完整路径的尾部){        更新某个全局变量的值    }    if (跳出循环的临界条件){        return;    }    对所有可能出现的情况进行递归} 常见题型1: 代码实现: #include <stdio.h> ; ; // 物品减数, 背包容量,最大价值maxValu…

一.深度优先搜索 它的定义是:递归探索图,必要时要回溯,同时避免重复. 关于深度优先搜索的伪代码如下: 左边DFS-Visit(V, Adj.s)是只实现visit所有连接某个特定点(例如s)的其他点.右边是实现整张图的visit,即DFS(v, Adj).DFS-Visit是DFS的重要组成模块. 用上图右侧的实例图解释下运作过程: 先从a出发,DFS-Visit到b上. 递归到b上,从b出发,DFS-Visit到e上. 递归到e上,从e出发,DFS-Visit到d上. 递归到d上,从d出发,…