HDU 6162 Ch’s gift (线段树+树链剖分)

【HDU 6162 Ch’s gift (线段树+树链剖分)】的更多相关文章

HDU 6162 Ch’s gift (线段树+树链剖分)

题意:给定上一棵树,每个树的结点有一个权值,有 m 个询问,每次询问 s, t , a, b,问你从 s 到 t 这条路上,权值在 a 和 b 之间的和.(闭区间). 析:很明显的树链剖分,但是要用线段树来维护,首先先离线,然后按询问的 a 排序,每次把小于 a 的权值先更新上,然后再查询,这样就是区间求和了,算完小于a的,再算b的,最答案相减就好了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #inc…

HDU 6162 - Ch’s gift | 2017 ZJUT Multi-University Training 9

/* HDU 6162 - Ch’s gift [ LCA,线段树 ] | 2017 ZJUT Multi-University Training 9 题意: N节点的树,Q组询问每次询问s,t两节点之间的路径上点权值在[a,b]之间的点权总和分析: 求出每个询问的LCA,然后离线按dfs顺序更新树状数组,即某点处树状数组中存的值为其所有祖先节点的值每个点处对答案的贡献为: 当其为第 i 个 lca 时, ans[i] -= 2 * query(a,b) , 再特判该节点当其为第 i…

2017 Multi-University Training Contest - Team 9 1002&&HDU 6162 Ch’s gift【树链部分+线段树】

Ch’s gift Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1354 Accepted Submission(s): 496 Problem Description Mr. Cui is working off-campus and he misses his girl friend very much. After a w…

HDU 6162 Ch’s gift （树剖 + 离线线段树）

Ch’s gift Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 662 Accepted Submission(s): 229 Problem Description Mr. Cui is working off-campus and he misses his girl friend very much. After a wh…

HDU 6162 Ch's gift（树链剖分+线段树）

题意: 已知树上的每个节点的值和节点之间的关系建成了一棵树,现在查询节点u到节点v的最短路径上的节点值在l到r之间的节点值的和. 思路: 用树链剖分将树映射到线段树上,线段树上维护3个值,max,min和sum即可. 接下来就是一个简单的线段树上的查询. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<vector> #inclu…

HDU 6162 Ch’s gift

Mr. Cui is working off-campus and he misses his girl friend very much. After a whole night tossing and turning, he decides to get to his girl friend's city and of course, with well-chosen gifts. He knows neither too low the price could a gift be sinc…