这应该属于一个比较麻烦的数据结构处理树上问题. 题目大意 给出一颗根节点编号为 \(1\) 的树,对于一个节点修改时在它的子树中对于深度奇偶性相同的节点加上这个权值,不同则减去这个值,单点查询. 分析 先给出这个问题的弱化版: 给出一颗根节点编号为 \(1\) 的树,对于一个节点修改时在它的子树中节点也加上这个权值,单点查询. 对于上面这个问题就好处理很多了,DFS序有一个性质,又称括号定理,在树中这个就会体现在再DFS序中以 \(a\) 为根节点的子树中的所有节点都在 \(a\) 的后面,且连…

\(tag\)没开够\(WA\)了一发... 求出\(dfs\)序,然后按深度分类更新与查询. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define R(a,b,c) for(register int a = (b); a <= (c); ++ a) #define nR(a,b,c) fo…

C. Propagating tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/383/problem/C Description Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numb…

题目链接:点击传送 E. Propagating tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consi…

C. Propagating tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n n…

Propagating tree Time Limit: 2000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 383C64-bit integer IO format: %I64d      Java class name: (Any)   Iahub likes trees very much. Recently he discovered an interesting tr…

C. Propagating tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/383/problem/C Description Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numb…

C. Propagating tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n n…

这道题明明没有省选难度啊,为什么就成紫题了QAQ 另:在CF上A了但是洛谷Remote Judge玄学爆零. 思路是DFS序+线段树. 首先这道题直观上可以对于每一次修改用DFS暴力O(n),然后对于询问O(1)解决. 但是这个方法实在是太耗时间了,因此我们想到了dfs序. 所谓dfs序,就是按照dfs(这里我们用先序遍历)的顺序给这颗树打上一个标签. 然后我们就可以把这颗树“拍平”,用一些支持区间修改单点查询的数据结构log级别解决问题了. 当然这样粗略地说一遍肯定会有人看不懂,还是通过一个实…

http://codeforces.com/problemset/problem/383/C 题目就是说,  给一棵树,将一个节点的值+val, 那么它的子节点都会-val, 子节点的子节点+val........这样类推, 给一系列操作,2是查询一个节点的值, 1是将一个节点的值+val. 首先,<dfs一遍求出dfs序以及每个点的深度, 然后建两颗线段树, 深度为奇数的建一棵, 偶数建一棵. 改变一个节点的值, 就把与他奇偶相同的那颗树全都+val, 不同的全都-val. 具体看代码>..…