树链剖分裸题. . .. 又要扩栈又要输入挂还卡格式....真无语 Tree Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1538    Accepted Submission(s): 261 Problem Description You are given a tree (an acyclic undirected conne…

HDU 5044 Tree field=problem&key=2014+ACM%2FICPC+Asia+Regional+Shanghai+Online&source=1&searchmode=source" target="_blank" style="">题目链接 就简单的树链剖分,只是坑要加输入外挂,还要手动扩栈 代码: #include <cstdio> #include <cstring>…

Tree Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1861    Accepted Submission(s): 545 Problem Description There are N (2<=N<=600) cities,each has a value of happiness,we consider two cities…

四边形优化DP Tree Construction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 868    Accepted Submission(s): 470 Problem Description Consider a two-dimensional space with a set of points (xi, yi) t…

Tree Problem Description You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N There are N - 1 edges numbered from 1 to N - 1. Each node has a value and each edge has a value. The initial v…

Tree Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1954    Accepted Submission(s): 573 Problem Description There are N (2<=N<=600) cities,each has a value of happiness,we consider two cities…

题目链接 题意: 有n个点的一棵树.其中树上有m条已知的链,每条链有一个权值.从中选出任意个不相交的链使得链的权值和最大. 思路: 树形DP.设dp[i]表示i的子树下的最优权值和,sum[i]表示不考虑i点时子树的最优权值和,即(j是i的儿子),显然dp[i]>=sum[i].那么问题是考虑i点时dp[i]的值是多少,假设有一条链通过i,且端点a和b都在i的子树里,即LCA(a,b)=i,如果考虑加上这条链的权值,那么a->i, b->i的路上的点v都不能有链经过它们(题目要求链不相交…

题意:给一棵树,两种操作: ADD1: 给u-v路径上所有点加上值k, ADD2:给u-v路径上所有边加上k,初始值都为0,问最后每个点和每条边的值,输出. 解法:树链剖分可做,剖出来如果直接用线段树来区间更新的话会TLE,所以要换一种姿势,有一种树链剖分的经典姿势就是看做树状数组一样,每次加值的时候,比如u->v之间加一个值k,那么在u处+k,v+1处-k即可,然后扫一遍,每次把当前位置要做的操作做完,此时总共加的值就是当前处的值,扫一遍的时候维护的是前缀的和,也就是两个ans不清零. 这题卡…

题意:一棵树上两种操作,操作1,改变u到v的每一点的值增加k,操作2,改变u到v每一条边值增加k.最后结束时问,每一点和每一条边的值. 初始时,点和边的值都为0. 分析: 很显然要用树链剖分,将点和边分别划分成连续一段的编号,然后就是维护一段一段的值了,给它增加一个值,由于题目只需要输出最后结果,那么可以用树桩数组维护. 以边为例,对于l~r的每个值增加k,利用树桩数组v[i]表示第i个值与前一个值的差值,那么第k条边的值就是sum{v[i]| 1 <= i <= k},更新时则只要给v[l]…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=3516 题意: 大概就是给你个下凸包的左侧,然后让你用平行于坐标轴的线段构造一棵树,并且这棵树的总曼哈顿距离最短 题解: 很容易得到转移方程: $$dp[i][j]=min \{ dp[i][k-1]+dp[k][j] + dis(uni(i,k-1),uni(k,j))\}$$ 其中$dp[i][j]$表示从$i$到$j$的最优解,$dis(i,j)$表示$i$和$j$之间的曼哈顿距离,$uni(i…

题意: 给出一棵\(n(1 \leq n \leq 10^5)\)个节点的树,每条边和每个点都有一个权值,初始所有权值为0. 有两种操作: \(ADD1 \, u \, v \, k\):将路径\(u \to v\)上所有节点的权值都加上\(k\) \(ADD2 \, u \, v \, k\):将路径\(u \to v\)上所有边的权值都加上\(k\) 最后输出每个点和边的权值. 分析: 看到树上成段更新,第一反应就是树链剖分,然而这样的算法并不能通过这道题目的数据. 注意到这道题的特殊之处在…

2015-07-15 问题简述: 给出一组节点关系,判断由这些节点组成的图是否为一颗树. 树只有一个根节点,每个节点只有一条边指向它,没有环. 原题链接:http://poj.org/problem?id=1308 解题思路: 使用并查集判断是否只有一个根节点是很简单的——让并查集种祖先的父亲是他自己即可方便计算其数量,一旦祖先数量超过一,它就不是树: 也可使用并查集判断图是否有环——当两个即将要链接的节点都有相同的祖先时,这就产生了一个环. 源代码: /* OJ: HDOJ ID: forev…

[HDOJ 5379] Mahjong tree 往一颗树上标号 要求同一父亲节点的节点们标号连续 同一子树的节点们标号连续 问一共同拥有几种标法 画了一画 发现标号有二叉树的感觉 初始标号1~n 根结点1能够标1或n 否则其它情况无法让以下的子树满足各自连续而且该根的儿子节点都要连续 根结点下的节点平分其它标号 画一画能够发现 每一个根下最多有两颗子树 否则无法满足条件 而且两颗子树占领剩余标号的左右两边 中间夹的必须是叶子 这样才干满足该根下的儿子节点标号连续 若根下仅仅有一颗子树 相同能够…

题目:pid=4925">hdoj 4925 Apple tree 来源:2014 Multi-University Training Contest 6 题意:给出一个矩阵,然后每一个格子中的数是2^(相邻格子的个数),然后要求不能取相邻的数,让取得数最大. 分析:这个题目有两种解法,一共是通解.网络流,还有一种是找规律,因为题目中数据是有规律的,所以能够找规律.非常多人是这样做的. 以下给出网络流的解法,事实上就是一个方格取数问题. 就是hdoj 1569 点击打开链接 的版本号,仅仅…

three version are provided. disjoint set, linked list version with weighted-union heuristic, rooted tree version with rank by union and path compression, and a minor but substantial optimization for path compression version FindSet to avoid redundanc…

DP/四边形不等式 这题跟石子合并有点像…… dp[i][j]为将第 i 个点开始的 j 个点合并的最小代价. 易知有 dp[i][j]=min{dp[i][j] , dp[i][k-i+1]+dp[k+1][j-(k-i+1)]+w(i,k,j)} (这个地方一开始写错了……) 即,将一棵树从k处断开成(i,k)和(k+1,i+j-1)两棵树,再加上将两棵树连起来的两条树枝的长度w(i,k,j) 其中,$ w(i,k,j)=x[k+1]-x[i]+y[k]-y[i+j-1] $ 那么根据四边形…

挺有意思的一道题,思路肯定是将图转化为Trie树,这样可以求得字典序.然后,按照trie的层次求解.一直wa的原因在于将树转化为线性数据结构时要从原树遍历,从trie遍历就会wa.不同结点可能映射为trie上的同一结点,如1->2 (a) 1->3(a) 2->4(b), 这是trie的结构是RT->a->b.然而,从结点3不能找到权重为b的路径.用RMQ求满足边界的rank最大值,通过sa找到该最大值对应的trie上的根.从而求解. /* 4601 */ #include…

并查集.需要考虑入度. #include <stdio.h> #include <string.h> #define MAXNUM 10005 int bin[MAXNUM]; int degree[MAXNUM]; int nums[MAXNUM]; int find(int x) { int r = x; while (bin[r] != r) r = bin[r]; return r; } int main() { ; int i, flg; ) { scanf("…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2915    Accepted Submission(s): 931 Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to…

Discription There is a skyscraping tree standing on the playground of Nanjing University of Science and Technology. On each branch of the tree is an integer (The tree can be treated as a connected graph with N vertices, while each branch can be treat…

CRB and Tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 690    Accepted Submission(s): 221 Problem Description CRB has a tree, whose vertices are labeled by 1, 2, -, N. They are connected…

Box Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2374    Accepted Submission(s): 718 Problem Description There are N boxes on the ground, which are labeled by numbers from 1 to N. The boxes…

版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/shengweisong/article/details/34099151 做了一上午,最终ac了 wa了一次主要是忘了还有环!.!   主要是运用并查集知识.又复习了一次!! 思路:输入之后找能不能成环,成环就不是,其次还要推断是不是有两个父节点,假设有两个父节点也不是.之后就找相关的祖先就好了. 还要注意:假设仅仅有一个节点,也是树.假设有两个或多个根节点也不是树:假设没有根节点也不是 链接pi…

文件可以重名.先按字典序将路径排序,再过滤掉公共前缀.其中的问题是'\'的ASCII比[A-Z0-9]大,将它替换为空格.否则字典序有问题. /* 1504 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include &l…

题目传送门 题意:给一棵树,问f (u, v) 意思是u到v的所有路径的边权值的异或和,问f (u, v) == s 的u,v有几对 异或+构造:首先计算f (1, u) 的值,那么f (u, v) == f (1, u) ^ f (1, v),f (u, v) == s -> f (1, u) == s ^ f (1, v), 异或的性质,再考虑s == 0特殊情况就好了 ;;; ; ; ; );        ; ;            ; ) ;                ;    …

神一样的题目.简言之,利用手段获得测试用例的第一行,输出结果.很显然利用wa, TLE, OLE等judge status可以获得测试用例.因此,果断Python写一个acm提交机器人.依赖lxml库.测试用例是The Ascii of H. #!/usr/env python import urllib2 import urllib import cookielib import logging from urlparse import * from time import sleep fro…

Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do some research on Spanning Tree. So Coach Pang decides to solve the following problem:Consider a bidirectional graph G with N vertices and M edges. All…

HDOJ 1501 Zipper [简单DP] Problem Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in i…

HDOJ 1501 Zipper [DP][DFS+剪枝] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10886 Accepted Submission(s): 3925 Problem Description Given three strings, you are to determine whether the third str…

Problem Description Consider a un-rooted tree T which is not the biological significance of tree or plant, but a tree as an undirected graph in graph theory with n nodes, labelled from 1 to n. If you cannot understand the concept of a tree here, plea…