As you might already know, Ada the Ladybug is a farmer. She grows a big fruit tree (with root in 0). There is a fruit on every node of the tree. Ada is competing in grafting competition and this is her masterpiece. The most valuable tree wins the com…

[题解]ADAGRAFT - Ada and Graft [SP33331] 传送门:\(\text{Ada and Graft}\) \(\text{[SP33331]}\) [题目描述] 给出一颗 \(n\) 个节点的树(根为 \(0\)),树的价值定义为每个节点的价值乘积.每个节点的值是其子树中不同颜色种类的数量. 给出每个节点上的颜色,求树价值.答案对 \(10^9+7\) 取模. [输入] 第一行包含一个整数 \(n\). 第二行包含 \(n-1\) 个整数 \(p_{i}\),分别表…

8.2.1.4 Index Merge Optimization 索引合并优化: 索引合并方法是用于检索记录 使用多个 范围扫描和合并它们的结果集到一起 mysql> show index from ClientInvestOrder; +-------------------+------------+------------------------+--------------+-------------+-----------+-------------+----------+------…

/* * This is an example build file that demonstrates how to use the build system for * require.js. * * THIS BUILD FILE WILL NOT WORK. It is referencing paths that probably * do not exist on your machine. Just use it as a guide. * * */   ({     //The…

题意 小 \(C\) 有一棵 \(n\) 个结点的有根树,根是 \(1\) 号结点,且每个结点最多有两个子结点. 定义结点 \(x\) 的权值为: 1.若 \(x\) 没有子结点,那么它的权值会在输入里给出,保证这类点中每个结点的权值互不相同. 2.若 \(x\) 有子结点,那么它的权值有 \(p_x\) 的概率是它的子结点的权值的最大值,有 \(1-p_x\) 的概率是它的子结点的权值的最小值. 现在小 \(C\) 想知道,假设 \(1\) 号结点的权值有 \(m\) 种可能性,权值第 \(i…

线段树优化dp的常见套路题,就是先按某个参数排序,然后按这个下标建立线段树,再去优化dp 本题由于要维护两个数据:最小值和对应的方案数,所以用线段树区间合并 /* dp[i]表示第i个套娃作为最内层的最小浪费空间 dp[i]=min(dp[j])+out[i]-in[i]; 那么按照out排序后按下标建立线段树,节点维护的是二元组(区间最小值,这个最小值对应的方案数) 求dp[i]时二分找到out[j]<=in[i]的区间[1,pos],然后线段树里查询再更新 处理方案数,线段树向上合并时,只保…

You are given a list of cities. Each direct connection between two cities has its transportation cost (an integer bigger than 0). The goal is to find the paths of minimum cost between pairs of cities. Assume that the cost of each path (which is the s…

暴力建图显然就是S->i连1,i->j'连inf(i为第j个力度能弹出的音符),j'->T连T[j]. 由于是“某棵子树中权值在某区间内的所有点”都向某个力度连边,于是线段树优化建图.由于是在树上所以需要可持久化线段树合并. 理论上可能空间会被卡,但是实际上并不能卡掉,边数最大点都不超过100W. 相比之下不太清楚为什么网上的dsu on tree做法为什么理论上就能过(可能是常数问题?),以及不理解为什么不用普通的启发式合并而非要用轻重链剖分. #include<cstdio&g…

题目链接:  51nod1674 题意:规定一个区间的价值为这个区间中所有数and起来的值与这个区间所有数or起来的值的乘积.现在l有一个 N 个数的序列,问所有n*(n+1)/2个区间的贡献的和对1000000007取模后的结果. 解法:暴力是O(n^2),我是尽量找了相差1的区间之间的规律,枚举区间的右端点可以发现当区间 [ l , r ]在右边新添一个数 r+1 时,答案要加上 [r+1,r+1].[r,r+1].[r-1,r+1] ...... [1,r+1].就可以每次存储 f [ i…

It is Dandiya Night! A certain way how dandiya is played is described: There are N pairs of people playing at a time. Both the person in a pair are playing Dandiya with each other. Since a person might get bored with the same partner, he can swap wit…