Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13865   Accepted: 5880 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or m…

http://poj.org/problem? id=1655 Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9072   Accepted: 3765 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a f…

题目链接 Balancing Act 就是求一棵树的重心,然后统计答案. #include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define for_edge(i,x) for(int i = H[x]; i; i = X[i]) const int INF = 1 << 30; const int N = 100000 + 10; int H[N…

题目链接: Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11845   Accepted: 4993 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of on…

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14247   Accepted: 6026 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or m…

题目大意:一棵n个节点的树,找出最大子树最小的节点. 题目分析:过程类似求重心. 代码如下: # include<iostream> # include<cstdio> # include<cstring> # include<vector> # include<queue> # include<list> # include<set> # include<map> # include<string>…

思路:DP求树的重心.对于每个结点$i$,$d_i=\displaystyle{\sum_{j\in s(i)}}d_j+1$.删去这个点能得到的最大子树大小即为$\max(\max\limits_{j\in s(i)}\{d(j)\},n-d(i))$.然后取最小的结点即可. #include<cstdio> #include<cctype> #include<vector> #include<cstring> inline int getint() {…

Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created…

树的重心即树上某结点,删除该结点后形成的森林中包含结点最多的树的结点数最少. 一个DFS就OK了.. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 222222 struct Edge{ int u,v,next; }edge[MAXN<<]; int NE,head[MAXN]; void addEdge(int u,int…

Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting…

求树的重心的模板题,size[u]维护以u为根的子树大小,f[u]表示去掉u后的最大子树. 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 using namespace std; 5 const int INF=0x7f7f7f7f; 6 const int N=20005; 7 int head[N],to[N*2],nxt[N*2],f[N],size[N]; 8 int tot,n,T,…

BZOJ1131: [POI2008]Sta 题意:给出一个N个点的树,找出一个点来,以这个点为根的树时,所有点的深度之和最大. 题解:记录每个点的深度,再根据根节点的深度和逐层推导出其他点的深度和. 我用的是BFS #include <stdio.h> #include <string.h> #include <iostream> #include <queue> using namespace std; typedef long long ll; int…

poj1010--邮票问题 DFSpoj1011--Sticks dfs + 剪枝poj1020--拼蛋糕poj1054--The Troublesome Frogpoj1062--昂贵的聘礼poj1077--Eightpoj1084--Square Destroyerpoj1085--Triangle War(博弈,極大極小搜索+alpha_beta剪枝)poj1088--滑雪poj1129--Channel Allocation 着色问题 dfspoj1154--letters (dfs)p…

http://poj.org/problem?id=1655 (题目链接) 题意 给出一棵树,求树上一点使得使得删除一点后该树的最大子树最小. solution 树的重心裸题. 随意取一点作为根节点,dfs维护当前节点的最大子树大小以及它的父亲子树大小,更新答案. 代码 // poj1655 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include&…

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12703   Accepted: 5403 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or m…

题意与分析 树的重心板子题. 值得考虑的是,重心究竟有哪些优秀的性质? 这里是一些网上能看到的性质: (判定性质)找到一个点,其所有的子树中最大的子树节点数最少(子树可以"倒着看"),那么这个点就是这棵树的重心. 以这个点为根,那么所有的子树(不算整个树自身)的大小都不超过整个树大小的一半. 树中所有点到某个点的距离和中,到重心的距离和是最小的:如果有两个重心,那么他们的距离和一样. 把两个树通过一条边相连得到一个新的树,那么新的树的重心在连接原来两个树的重心的路径上. 把一个树添加或…

Load Balancing Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83008#problem/H Description The infamous University of Kala Jadu (UKJ) have been operating underground for the last fourteen centuries t…

树的重心 我们先来认识一下树的重心. 树的重心也叫树的质心.找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后,生成的多棵树尽可能平衡. 根据树的重心的定义,我们可以通过树形DP来求解树的重心. 设\(Max_i\)代表删去i节点后树中剩下子树中节点最多的一个子树的节点数.由于删去节点i至少将原树分为两部分,所以满足\(\ \frac{1}{2}n \leq Max_i\),我们要求的就是一个\(i\),使得\(Max_i\)最小. 对于Max数组,我们可以列…

Method for balancing a binary search tree. A computer implemented method for balancing a binary search tree includes locating a node in a binary search tree, determining whether a depth of the located node is greater than a threshold, and performing…

Atitit  图像处理Depixelizing Pixel Art像素风格画的矢量化 在去年的时候,偶然看到hqx算法. 一个高质量的插值放大算法. 与双线性插值等插值算法相比,这个算法放大后对人眼保护相对比较好. 没有双线性插值看起来模糊,固然,也抽空把算法简单优化了一下. 常规的图像填充采样放大 "经典"的图像填充采样放大,应用线性滤波器,或导出于分析插值理论.或导出于信号处理理论.使用滤波器的实例,比如"最近邻,Bicubic和Lancosz [ Wolberg 19…

Gerald is into Art Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought a special board to attach it to the wall and place the paintings on the board. The board has shape of an a…

http://blogs.mathworks.com/loren/2007/03/01/creating-sparse-finite-element-matrices-in-matlab/ Loren on the Art of MATLAB March 1st, 2007 Creating Sparse Finite-Element Matrices in MATLAB I'm pleased to introduce Tim Davis as this week's guest blogge…

这几天在做一个项目时需要在Android中使用OSGi框架(Apache Felix),于是在一个android 4.4.2 版本系统的某品牌的平板上实验. 实验内容很简单:把felix包里的felix.jar包和一些bundles的jar包用android sdk里的dx及aapt工具转化为包含dex字节码的jar 包.这样使这些jar包能在Android上跑起来.(因为Android上使用的是在Dalvik虚拟机而不是标准的java虚拟机,所以"原生"的 jar包不能直接在Andr…

原文地址: https://github.com/ZhaoKaiQiang/AndroidDifficultAnalysis/blob/master/10.ART%E3%80%81JIT%E3%80%81AOT%E3%80%81Dalvik%E4%B9%8B%E9%97%B4%E6%9C%89%E4%BB%80%E4%B9%88%E5%85%B3%E7%B3%BB%EF%BC%9F.md ART.JIT.AOT.Dalvik之间有什么关系? JIT与Dalvik JIT是"Just In Tim…

How To Use HAProxy to Set Up MySQL Load Balancing Dec  2, 2013 MySQL, Scaling, Server Optimization Ubuntu, Debian         Prelude HAProxy is an open source software which can load balance HTTP and TCP servers. In the previous article on HAProxy we co…

Balancing Act   Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the…

题目链接:http://poj.org/problem?id=3107 求树的重心,所谓树的重心就是:在无根树转换为有根树的过程中,去掉根节点之后,剩下的树的最大结点最小,该点即为重心. 剩下的数的 最大结点dp[i]=max(max(s[j]),n-s[i])  其中的s[i]为以i为根节点的子树的结点总数,j为i的孩子. //思路和代码都是比较好理解的 代码: #include<iostream> #include<cstdio> #include<algorithm&g…

C. Load Balancing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In the school computer room there are n servers which are responsible for processing several computing tasks. You know the…

与 iOS 相比,Android 的用户体验有个相对糟糕的开始.在很长的时间里,界面一直丑小鸭,卡顿也是挥不去的痛.不过,在 Google 的全力推动,以及硬件厂商的响应下,Android 还是跨越各种阻碍,逐渐壮大起来了. 在此过程中,Google 也在经历着重大的变化.它逐渐从一个只重视数据的公司,转变为一个重视设计和用户体验的公司.从 Android 4.0 开始,Android 拥有了自己的设计语言和应用设计指导.与此同时,Google 也在着手解决卡顿问题.Android 4.1 的“…

B. Gerald is into Art Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/560/B Description Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought…