Hdu 3966-Aragorn's Story LCT,动态树】的更多相关文章

HDU Aragorn's Story 题目链接 树抛入门裸题,这题是区间改动单点查询,于是套树状数组就OK了 代码: #include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespace std; const int N = 50005; inline int lowbit(int x) {return x&(-x);} int dep…
Query on The Trees Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 4091    Accepted Submission(s): 1774 Problem Description We have met so many problems on the tree, so today we will have a que…
HDU 3966 Aragorn's Story 先把树剖成链,然后用树状数组维护: 讲真,研究了好久,还是没明白 树状数组这样实现"区间更新+单点查询"的原理... 神奇... #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #inc…
pid=3966" target="_blank" style="">题目链接:hdu 3966 Aragorn's Story 题目大意:给定一个棵树,然后三种操作 Q x:查询节点x的值 I x y w:节点x到y这条路径上全部节点的值添加w D x y w:节点x到y这条路径上全部节点的值降低w 解题思路:树链剖分,用树状数组维护每一个节点的值. #pragma comment(linker, "/STACK:1024000000,1…
HDU - 3966 Aragorn's Story Time Limit: 3000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of ene…
题目链接: Hdu 3966 Aragorn's Story 题目描述: 给出一个树,每个节点都有一个权值,有三种操作: 1:( I, i, j, x ) 从i到j的路径上经过的节点全部都加上x: 2:( D, i, j, x ) 从i到j的路径上经过的节点全部都减去x: 3:(Q, x) 查询节点x的权值为多少? 解题思路: 可以用树链剖分对节点进行hash,然后用线段树维护(修改,查询),数据范围比较大,要对线段树进行区间更新 #include <cstdio> #include <…
Aragorn's Story Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies w…
题目:http://acm.hdu.edu.cn/showproblem.php?pid=3966 Aragorn's Story Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7658    Accepted Submission(s): 2024 Problem Description Our protagonist is the…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3966 题意:给一棵树,并给定各个点权的值,然后有3种操作: I C1 C2 K: 把C1与C2的路径上的所有点权值加上K D C1 C2 K:把C1与C2的路径上的所有点权值减去K Q C:查询节点编号为C的权值 题解:就是树链剖分具体看代码,还有注释. #include <iostream> #include <cstring> using namespace std; const…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4010 题意; 先给你一棵树,有 \(4\) 种操作: 1.如果 \(x\) 和 \(y\) 不在同一棵树上则在\(x-y\)连边. 2.如果 \(x\) 和 \(y\) 在同一棵树上并且 \(x!=y\) 则把 \(x\) 换为树根并把 \(y\) 和 \(y\) 的父亲分离. 3.如果 \(x\) 和 \(y\) 在同一棵树上则 \(x\) 到 \(y\) 的路径上所有的点权值\(+w\). 4…