Fling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 354    Accepted Submission(s): 143 Problem Description Fling is a kind of puzzle games available on phone.This game is played on a board wit…

HDU.5692 Snacks ( DFS序 线段树维护最大值 ) 题意分析 给出一颗树,节点标号为0-n,每个节点有一定权值,并且规定0号为根节点.有两种操作:操作一为询问,给出一个节点x,求从0号节点开始到x节点,所能经过的路径的权值最大为多少:操作二为修改,给出一个节点x和值val,将x的权值改为val. 可以看出是树上修改问题.考虑的解题方式有DFS序+线段树,树链剖分,CXTree.由于后两种目前还不会,选择用DFS序来解决. 首先对树求DFS序,在求解过程当中,顺便求解树上前缀和(p…

Fling Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 455    Accepted Submission(s): 190 Problem Description Fling is a kind of puzzle games available on phone.This game is played on a board wit…

Necklace/center> 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5727 Description SJX has 2*N magic gems. N of them have Yin energy inside while others have Yang energy. SJX wants to make a necklace with these magic gems for his beloved BHB. To avoid…

Cannon Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4499 Description In Chinese Chess, there is one kind of powerful chessmen called Cannon. It can move horizontally or vertically along the chess grid. At eac…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1175 解题思路:从出发点开始DFS.出发点与终点中间只能通过0相连,或者直接相连,判断能否找出这样的路径. #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define N 1…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=5547 题目: Sudoku Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2372    Accepted Submission(s): 804 Problem Description   Yi Sima was one of the be…

题目链接: F - Auxiliary Set HDU - 5927 学习网址:https://blog.csdn.net/yiqzq/article/details/81952369题目大意一棵节点数为n的有根数,根节点为1,一开始所有的点都是重点,接下来有q次询问,每次询问把m个点变为轻点,问你树中还有多少个重点. 重点应该满足的条件为: 1.它本身是重点. 2.它为两个重点的最近公共祖先. 每次询问之后在下次询问前,所有的点都恢复为重点. 具体思路:对于每个点保存他的深度.因为每次输入的数…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4714 本来想直接求树的直径,再得出答案,后来发现是错的. 思路:任选一个点进行DFS,对于一棵以点u为根节点的子树来说,如果它的分支数大于1,那么我们把这颗子树从整棵树上剪下来(优先减去),同时把这颗子树的分支留下两个,其它多余的也剪掉,然后把剪下来的这些部分连接到根节点那里,从而形成一条直链,总代价就是我们减的次数+把剪下来的部分连接到根节点+把最后的直链连成环.在这里剪的次数=把剪下来的部分连接…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6228 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Problem DescriptionConsider a un-rooted tree T which is not the biological significance of tree or plant, but a tre…