Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1484    Accepted Submission(s): 347 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…

树链剖分第一题QAQ,纪念下 #pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef long long ll; const ll mod = 1e9 + 7; const int maxn = 1e5 + 10; #define…

List wants to travel Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 429    Accepted Submission(s): 92 Problem Description A boy named List who is perfect in English. Now he wants to travel an…

Yaoge’s maximum profit Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 982    Accepted Submission(s): 274 Problem Description Yaoge likes to eat chicken chops late at night. Yaoge has eaten too…

Little Devil I Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 894    Accepted Submission(s): 296 Problem Description There is an old country and the king fell in love with a devil. The devil…

Problem Aragorn's Story (HDU 3966) 题目大意 给定一颗树,有点权. 要求支持两种操作,将一条路径上的所有点权值增加或减少ai,询问某点的权值. 解题分析 树链剖分模板题. 实质上树链剖分进行了点对点的一次映射,保证了重链上的点在线段树上的位置是连续的. 树链剖分的两个性质(转): 性质1:如果(v,u)为轻边,则siz[u] * 2 < siz[v]: 性质2:从根到某一点的路径上轻链.重链的个数都不大于logn. 保证了一个区间的时间复杂度是log2(n).…

传送门:Problem 3966 https://www.cnblogs.com/violet-acmer/p/9711441.html 学习资料: [1]线段树区间更新:https://blog.csdn.net/zhhe0101/article/details/53871453  https://yq.aliyun.com/articles/252586 [2]树链剖分:http://blog.sina.com.cn/s/blog_7a1746820100wp67.html    https…

题意: http://acm.hdu.edu.cn/showproblem.php?pid=3966 给一棵树,并给定各个点权的值,然后有3种操作: I x y z : 把x到y的路径上的所有点权值加上z D x y z:把x到y的路径上的所有点权值减去z Q z:查询节点编号为x的权值 这里主要放下用树状数组维护的模板 区间修改单点查询 好像用线段树更好? em.... 两种都放好了~ 好像说hduoj是windows系统容易爆栈 手动扩栈加这句 #pragma comment(linker,…

思路:这个树链剖分其实还是比较明显的.将边按权值排序后插入线段树,然后用线段树查找区间中比某个数小的数和,以及这样的数的个数.当A<=B时,就全部建新的管子. 对于A>B的情况比较 建一条由S->T的管子后将这根管子容量扩到最大能得到的容量  与   将所有预算都用来扩大管子容量不建新管子得到的最大容量 做比较 ,选最大的. 扩容量能得到的最大权值可以同过二分枚举答案,用树链剖分判断. #include<set> #include<map> #include<…

思路:树链剖分入门题,我这门入得好苦啊,程序很快写出来了,可是在LCA过程中把update函数里的左右边界位置写反了,一直RE到死. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<set> #include<map> #include<cmath> #include<queue> #include<cstdio> #include<vec…