D. Acyclic Organic Compounds   You are given a tree T with n vertices (numbered 1 through n) and a letter in each vertex. The tree is rooted at vertex 1. Let's look at the subtree Tv of some vertex v. It is possible to read a string along each simple…

[题目]D. Acyclic Organic Compounds [题意]给定一棵带点权树,每个点有一个字符,定义一个结点的字符串数为往下延伸能得到的不重复字符串数,求min(点权+字符串数),n<=300000,time=3s. [算法]trie合并||hash+线段树合并||dsu on tree [题解]维护每个节点的Trie,那么每个节点的不重复字符串数是Trie的节点数. 每个节点Tire的根设为这个节点的字符(不是空字符). 这样Trie的合并就很方便了,merge(a,b)表示将b…

trie合并的裸题...因为最多只有n个点,所以最多合并n次,复杂度$O(N*26)$. #include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> using namespace std; , inf=1e9; ]; ], size;}tree[maxn]; int n, x, y, ans, cnt, tot; int last[maxn], c[maxn]; char…

题意: 给一以1为根的字符树,给出每个节点的字符与权值,记 $diff_{x}$ 为从 $x$ 出发向下走,能走到多少不同的字符串,求问最大的$diff_{x} + c_{x}$,并求有多少个 $diff_{x} + c_{x}$. 解法: 考虑$dfs$,从下到上启发式合并 $Trie$ 树,效率 $O(nlogn)$. #include <iostream> #include <cstdio> #include <cstring> #define N 300010…

给n<=300000的树,每个点上有一个字母,一个点的权值为:从该点出发向下走到任意节点停下形成的不同字符串的数量,问最大权值. 题目本身还有一些奇怪要求在此忽略.. Trie合并的模板题. #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> //#include<queue> #include<algorithm> #include&…

A.The Two Routes(BFS) 给出n个城镇,有m条铁路,铁路的补图是公路,汽车和火车同时从1出发,通过每条路的时间为1,不能同时到达除了1和n的其它点,问他们到达n点最少要用多长时间. 因为是补图,那么一定有一条路是可以直接从1到达n的.那么我们把剩下的用bfs求一下即可. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # incl…

1. CF 438D The Child and Sequence 大意: n元素序列, m个操作: 1,询问区间和. 2,区间对m取模. 3,单点修改 维护最大值, 取模时暴力对所有>m的数取模. 因为取模后至少减半, 复杂度$O(nlognlogC)$ 2. CF 431E Chemistry Experiment 大意: n个试管, 第$i$个试管有$a_i$单位水银, m个操作: 1, 修改$a_x$改为$v$. 2, 将$v$单位水倒入试管, 求一种方案使得有水的试管水银与水总量的最大…

常见的coupled models phase English paper WRF-Chem mechanism public data 一些重要的结论 干空气的状态方程 ECWMF驱动WRF 常见的coupled models RegCM4: Regional Climate Model version 4 GATOR-GCMOM: the Gas, Aerosol, Transport, Radiation, General-Circulation, Mesoscale, and Ocean…

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Given a Weighted Directed Acyclic Graph and a source vertex in the graph, find the shortest paths from given source to all other vertices. For a general weighted graph, we can calculate single source shortest distances in O(VE) time using Bellman–For…