HDU 5052 LCT

【HDU 5052 LCT】的更多相关文章

Yaoge's maximum profit Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 516 Accepted Submission(s): 150 Problem Description Yaoge likes to eat chicken chops late at night. Yaoge has eaten to…

hdu 5052 树链剖分

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…

Yaoge’s maximum profit HDU - 5052

Yaoge’s maximum profit Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5052 Problem Description Yaoge likes to eat chicken chops late at night. Yaoge has eaten too ma…

Hdu 2475-Box LCT,动态树

Box Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2813 Accepted Submission(s): 821 Problem Description There are N boxes on the ground, which are labeled by numbers from 1 to N. The boxes…

hdu 4010 Lct动态链接树

#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <memory> #include <cstdlib> #include <cstring> #include <cmath> #include <vector> #include <cassert> #include <string…

HDU 5052 Yaoge’s maximum profit 光秃秃的树链拆分 2014 ACM/ICPC Asia Regional Shanghai Online

意甲冠军: 特定n小点的树权. 以下n每一行给出了正确的一点点来表达一个销售点每只鸡价格的格以下n-1行给出了树的侧以下Q操作 Q行 u, v, val 从u走v,程中能够买一个鸡腿,然后到后面卖掉,输出max(0, 最大的收益) 然后给[u,v]路径上点点权+=val 思路: 树链剖分裸题.记录区间的最大最小值,→走的答案和←走的答案. 官方题解:点击打开链接 #pragma comment(linker, "/STACK:1024000000,1024000000") #inc…

Hdu 5052 Yaoge’s maximum profit（树链剖分）

题目大意: 给出一棵树.每一个点有商店.每一个商店都有一个价格,Yaoge每次从x走到y都能够在一个倒卖商品,从中得取利益.当然,买一顶要在卖之前.可是没次走过一条路,这条路上的全部商品都会添加一个v. 输出每次的最大利益. 思路分析: 非常easy想到树链剖分,但是关键在于怎样维护这样一个变量.使得每次都要让买的再卖的前面. 维护变量 ltr 和 rtl .表示从左去右和从右去左. 剖分熟练的时候.推断x 和 y的深度,一步一步往上爬. 然后维护区间最大值和最小值.爬的时候更新答案. ...4…