http://acm.hdu.edu.cn/showproblem.php?pid=5102 给一棵树,求出所有节点的距离中前k小的路径长度和 由于路径长度的定义为两点之间的边的个数,所有遍历1~n-1条边组成的路径,暴力擦线过,3000+ms #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include &l…

一棵树,不一定是二叉树,在每个结点最多只属于一条链的情况下,处理出其中最长的前k个的长度. 最近训练赛做到两道题了,有必要总结一下. 不过我不知道是否有更专门的叫法. 借鉴了这位大佬的博客:https://www.cnblogs.com/Aragaki/p/11754534.html 例题1. 2019-2020 ACM-ICPC Brazil Subregional Programming Contest D - Denouncing Mafia https://codeforces.com/…

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…

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…

Group Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1959    Accepted Submission(s): 1006 Problem Description There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i…

Rabbit Kingdom Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1999    Accepted Submission(s): 689 Problem Description Long long ago, there was an ancient rabbit kingdom in the forest. Every ra…

A*算法是一类贪心算法,其可以用于寻找最优路径.我们可以利用A*算法来求第k短路径. 一条路径可以由两部分组成,第一部分是一个从出发到达任意点的任意路径,而第二部分是从第一部分的末端出发,到终点的最短路径.两部分正好可以组成一条路径,且每一条路径都可以分解这两部分(允许任意一部分为空).因此当我们已知第一部分的路径A时,设第二部分为B,我们可以尝试预估完整的路径A+B的费用(距离),我们将公式定义为:f(A)=g(A)+h(A).其中g(A)表示第一部分A的已知长度,而h(A)表示路径A到终点的…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2852 题目大意:操作①:往盒子里放一个数.操作②:从盒子里扔掉一个数.操作③:查询盒子里大于a的第K小数. 解题思路: 由于模型是盒子,而不是序列,所以可以用树状数组的顺序维护+逆序数思想. 对应的树状数组Solution: 放一个数 $Add(val,1)$ 类似维护逆序数的方法,对应位置上计数+1. 注意Add的while范围要写成$while(x<maxn)$ 如果范围不是最大,那么会导致某些…