http://vjudge.net/problem/17662 loli蜜汁(面向高一)树形dp水题 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct nodeTreeDP { struct node {int nxt, to, w;} E[203]; int n, q, cnt, point[103], apple[103], left[103], ri…

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in average O(1) time and uses…

CJOJ 1976 二叉苹果树 / URAL 1018 Binary Apple Tree(树型动态规划) Description 有一棵苹果树,如果树枝有分叉,一定是分2叉(就是说没有只有1个儿子的结点)这棵树共有N个结点(叶子点或者树枝分叉点),编号为1-N,树根编号一定是1.我们用一根树枝两端连接的结点的编号来描述一根树枝的位置.现在这颗树枝条太多了,需要剪枝.但是一些树枝上长有苹果. 给定需要保留的树枝数量,求出最多能留住多少苹果.下面是一颗有 4 个树枝的树. 2 5 \ / 3 4…

Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by integers the root of binary apple tree, points of branch…

1018. Binary Apple Tree Time limit: 1.0 secondMemory limit: 64 MB Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enu…

Binary Apple Tree Time Limit: 1000ms Memory Limit: 16384KB This problem will be judged on Ural. Original ID: 101864-bit integer IO format: %lld      Java class name: (Any)     Let's imagine how apple tree looks in binary computer world. You're right,…

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in average O(1) time and uses…

Binary Search Tree Iterator Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in…

[题目链接]:http://www.lydsy.com/JudgeOnline/problem.php?id=1018 [题意] [题解] 按照这里的题解写的http://blog.csdn.net/popoqqq/article/details/44116729 主要思路就是根据最小单元的合并方式推出整个线段的合并. 其中要往左走和往右走的过程,可以一开始让那个点往左移动,看看最远能到哪里,右边的点也一样,一直向右移动看看最远能到哪; 然后在最左和最右之间求联通性; a[x][y]表示这个区间…

[题目链接] 点击打开链接 [算法] 树链剖分 对于线段树的每个节点,记录这段区间的最小值,最小值的个数,值为0的个数,此外,还要维护两个懒惰标记 [代码] 本题细节很多,写程序时要认真严谨! #include<bits/stdc++.h> using namespace std; #define MAXN 100010 #define MAXLOG 20 const int INF = 1e9; int i,n,m,tot,opt,u,v,c,x,y,timer,Lca,tmp; int d…