There are two ways to conduct BFS on tree. Solution 1 -- Given level Use recursion to find given level, and print. /*Function to print level order traversal of tree*/ printLevelorder(tree) for d = 1 to height(tree) printGivenLevel(tree, d); /*Functio…

You are given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id. For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. Th…

https://www.hackerrank.com/contests/illuminati/challenges/tree-covering 这道题先是在上次交流讨论了一下,然后两位百度的朋友先写完代码share出来了,觉得是道很好的题,就做了一下.https://gist.github.com/coder32167/6964331 https://gist.github.com/snakeDling/6965299基本思想是贪心.根据题意,所选的点必然是叶子节点,那么首先找出树的直径,直径上…

Question Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [1…

BFS与DFS常考算法整理 Preface BFS(Breath-First Search,广度优先搜索)与DFS(Depth-First Search,深度优先搜索)是两种针对树与图数据结构的遍历或搜索算法,在树与图相关算法的考察中是非常常见的两种解题思路. BFS与DFS常考算法整理 Definition of DFS and BFS How to Implement DFS and BFS DFS How to explore as far as possible How to backt…

菜鸟心得.... 不对请指出....... /*BFS,广度优先搜索树,用最简单的2叉树来举例, 树的结构如下: A B C D E F GH I J K L M N O广度优先搜索树, 顺序应该是ABCDEFGHIJKLMNO; 意思是,先搜索完 上一层的节点,再开始搜索下一层的节点;那么在BFS中, 会使用到一种数据结构----队列队列的特点是,先进先出, 就像一条水管一样,这头进,那头出.尾----------------------头->> F E D C B A ->>&…

Problem Link: http://oj.leetcode.com/problems/minimum-depth-of-binary-tree/ To find the minimum depth, we BFS from the root and record the depth. For each level we add 1 to the depth and return the depth value when we reach a leaf. The python code is…

A:树形DP 给出一棵树,但是它的边是有向边,选择一个城市,问最少调整多少条边的方向能使一个选中城市可以到达所有的点,输出最小的调整的边数,和对应的点 要改变的边的权值为1,不需要改变的边的权值为0, 两次dfs 第一次算出以1点为根节点到所有点要改变的边数,第二次以1为根节点向下遍历节点   算出每一个点到达所有点要改变的边数, dp[son]+=(dp[root]-dp[son])+((tree[i].val)?-1:1),某一点的值是他父节点的值减去他以前的值再考虑他与父节点之间的边的方向…

我就是来复习Dinic算法的,仅10天不写,我已经退化成写一遍+调试需要接近一个小时了,当然其中不乏在网上乱逛的时间… 赞成从S源点连一条单向边,反对向T汇点连一条单向边,朋友关系连双向边. 但是总感觉自己看到题目不能一下想到这是网络流,感觉这些题都是给一个图,求最优之类. program vote; type ptype=^node; node=record v,w,flow:longint; op,next:ptype; end; ; ..maxn] of ptype; visit:..ma…

最后两点怎么搞都要30s+,但是我不会什么优化啊…暂时就这样吧.Dinic的时间复杂度是O(N^2*M) 这题和TDL的幼儿园模板是一样的. 这次写网络流给自己计时了,大约是40min左右,后来都跑去倒腾后面两组数据去了… program profit; type ptype=^node; node=record v,w,flow:longint; next:ptype; end; +; inf=maxlongint ; ..maxn] of ptype; visit:..maxn] of bo…