题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on t…

题目大意: 有一棵结点个数为n的树,有m个操作,可以将一段路径上每条边的权值+1或询问某一个边的权值. 思路: 树链剖分+线段树. 轻重链划分本身比较简单,主要需要思考如何用线段树维护每条链. 当x,y不在同一条链上时,先处理深度大的链,对于每一个链,建立一棵动态开点的线段树,用一个数组len顺序记录每一条边在链中的编号,然后维护len[x]+1到len[top[x]]这一区间的权值即可. 处理轻边时,可以直接用一个数组保存它的权值. 因为轻重边肯定是交替的,因此每次循环都可以先维护一个重边,再…

洛谷 P3038 [USACO11DEC]牧草种植Grass Planting 洛谷传送门 JDOJ 2282: USACO 2011 Dec Gold 3.Grass Planting JDOJ传送门 Description Problem 3: Grass Planting [Travis Hance, 2011] Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roa…

Grass Planting 题意 给出一棵树,树有边权.每次给出节点 (u, v) ,有两种操作:1. 把 u 到 v 路径上所有边的权值加 1.2. 查询 u 到 v 的权值之和. 分析 如果这些值不是在树上,而是在区间上,那么凭借线段树.树状数组可以很轻松的解决,但是在树上则不能直接操作. 树链剖分就是将树上的节点映射到区间上,从而实现区间操作. 学习树链剖分前需要掌握的知识点:线段树.LCA. 参考blog 认真读完这篇 blog ,跟着算法流程走一遍差不多就懂了. code #incl…

题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on t…

题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on t…

题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on t…

题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on t…

题解:仍然是无脑树剖,要注意一下边权,然而这种没有初始边权的题目其实和点权也没什么区别了 代码如下: #include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define lson root<<1 #define rson root<<1|1 using namespace std; struct…

图很丑.明显的树链剖分,需要的操作只有区间修改和区间查询.不过这里是边权,我们怎么把它转成点权呢?对于E(u,v),我们选其深度大的节点,把边权扔给它.因为这是树,所以每个点只有一个父亲,所以每个边权都可以唯一地.不重复地转移到点上去(除了根节点).但是做区间操作时就要注意一下,区间两边端点的LCA的权值是不可以用的. 这么简单的模板题就直接放代码了: #include<iostream> #include<cstring> #include<cstdio> #defi…