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SP12005 GRASSPLA - Grass Planting 题解

题目传送门更好的阅读体验 Solution I 这几乎是一道树链剖分模板题，和模板题唯一的区别在于这题维护的是边权。因为除了根以外的节点都有父亲，但是叶子数量很多，所以我们让深度大的节点存储边的信息，就方便处理很多了。在操作的时候，因为深度大的节点存储的才是边的信息，所以最顶端的

「USACO11DEC」Grass Planting G 题解 (树链剖分)

题目简介给出一棵 \(N\) 个节点的树，有 \(M\) 个操作，操作为将一条路径上的边权加一或询问某条边的权值。分析点差分与边差分的区别是：点差分计入 \(lca\) ，边差分不计 \(lca\)。模板树链剖分是对点统计，类似点差分。本题是对边统计，只需要去掉 \(lca\) 的计算即可。 \(AC\ Code\)

【21天精听打卡 2/21】20211110 TED精听 SusanGraham: A new way to restore Earth’s biodiversity—from the air

20211110 TED精听 SusanGraham: A new way to restore Earth’s biodiversity—from the air Every year human change 10 million hm of plant, and not for the better. [Every year, humans change 10 million hectares of land, and not for the better.] Right now , the

LeetCode 1042. Flower Planting With No Adjacent

原题链接在这里：https://leetcode.com/problems/flower-planting-with-no-adjacent/ 题目： You have N gardens, labelled 1 to N. In each garden, you want to plant one of 4 types of flowers. paths[i] = [x, y] describes the existence of a bidirectional path from

leetcode1042 Flower Planting With No Adjacent

1 """ 2 You have N gardens, labelled 1 to N. In each garden, you want to plant one of 4 types of flowers. 3 paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y. 4 Also, there is no garden that ha

2019 牛客暑期多校第三场 F Planting Trees (单调队列+尺取）

题目：https://ac.nowcoder.com/acm/contest/883/F 题意：求一个矩阵最大面积，这个矩阵的要求是矩阵内最小值与最大值差值<=m 思路：首先我们仔细观察范围，我们就知道可以n^3,前面这题我（看付队博客）讲过求一个最大的什么矩阵就是分两种情况，第一种：枚举上下边界，转化为一维，复杂度n^3 第二种：枚

[USACO11DEC]牧草种植Grass Planting

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

[USACO11DEC]牧草种植Grass Planting

题目描述给出一棵n个节点的树，有m个操作，操作为将一条路径上的边权加一或询问某条边的权值。输入输出样例输入样例#1：复制 4 6 1 4 2 4 3 4 P 2 3 P 1 3 Q 3 4 P 1 4 Q 2 4 Q 1 4 输出样例#1：复制 2 1 2 【解题思路】【code】 1 // luogu-judger-enable-o2 2 #include <cst

Leetcode-5056 Flower Planting With No Adjacent(不邻接植花)

1 #define _for(i,a,b) for(int i = (a);i < b;i ++) 2 3 int n,m; 4 5 class Solution 6 { 7 public: 8 int color[10000]; 9 vector<vector<int>> a;10 bool OK(int t)11 {12 for(int j=0; j<a[t].si