The 2017 ACM-ICPC Asia Beijing Regional Contest

地址

Rank Solved A B C D E F G H I J
51/384 4/10 . . ? . O O . O . O

O: 当场通过

?: 赛后通过

.: 尚未通过

A Domains

unsolved


B K-Dimensional Foil

unsolved


C Graph

upsolved by chelly



chelly‘s solution

很显然的思路就是莫队+并查集

但众所周知并查集可以支持可撤销,但不是很好支持可持久化

于是就可以用上回滚莫队的套路了,回滚莫队可以把一般莫队的删除操作变成撤销操作,复杂度不改变

于是这题用回滚莫队+可撤销并查集即可解决

注意回滚莫队的时候,处理右半边的点的时候,不能连向左半边的点,否则rollback的时候跨越中间的边并没有撤销

D Chinese Checkers

unsolved


E Cats and Fish

solved by chelly



chelly‘s solution

F Secret Poems

solved by chelly



chelly‘s solution

G Liaoning Ship’s Voyage

unsolved


H Puzzle Game

solved by chelly



chelly‘s solution

I Colored Nodes

unsolved


J Pangu and Stones

solved by chelly



chelly‘s solution

Replay

原文地址:https://www.cnblogs.com/Amadeus/p/9986324.html

时间: 2024-11-09 14:43:19

The 2017 ACM-ICPC Asia Beijing Regional Contest的相关文章

hdu6206 Apple 2017 ACM/ICPC Asia Regional Qingdao Online

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6206 题目: Apple Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 530    Accepted Submission(s): 172 Problem Description Apple is Taotao's favouri

2017 ACM/ICPC Asia Regional Shenyang Online spfa+最长路

transaction transaction transaction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1496    Accepted Submission(s): 723 Problem Description Kelukin is a businessman. Every day, he travels arou

ACM ICPC Central Europe Regional Contest 2013 Jagiellonian University Kraków

ACM ICPC Central Europe Regional Contest 2013 Jagiellonian University Kraków Problem A: Rubik's RectangleProblem B: What does the fox say?Problem C: Magical GCDProblem D: SubwayProblem E: EscapeProblem F: DraughtsProblem G: History courseProblem H: C

2018-2019, ICPC, Asia Yokohama Regional Contest 2018 (Gym - 102082)

2018-2019, ICPC, Asia Yokohama Regional Contest 2018 A - Digits Are Not Just Characters 签到. B - Arithmetic Progressions 题意:从给定的集合中选出最多的数构成等差数列. 题解:数字排序后,设\(dp[i][j]\)表示等差数列最后一个数字为\(a[i]\),倒数第二个数字为\(a[j]\)的最大个数.然后对于每一位枚举 \(i\),\(lower\_bound()\)找有无合法的

2017 ACM/ICPC Asia Regional Beijing Online

A Visiting Peking University 直接模拟 B Reverse Suffix Array unsolved C Matrix 预处理每列前缀和以及每个每个行区间的每列的最小值,枚举行的所有区间,求最长连续字段和. D Agent Communication unsolved E Territorial Dispute n=1,2时不行 n=3时必须三点共线 n>3一定可以,任意找四个点判断即可. 好像也可以跑凸包,不在凸包上的点单独分开,如果全部在凸包上,找两个不相邻的点

HDU - 6215 2017 ACM/ICPC Asia Regional Qingdao Online J - Brute Force Sorting

Brute Force Sorting Time Limit: 1 Sec  Memory Limit: 128 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=6215 Description Beerus needs to sort an array of N integers. Algorithms are not Beerus's strength. Destruction is what he excels. He can destr

2017 ACM/ICPC Asia Regional Shenyang Online

cable cable cable Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2084    Accepted Submission(s): 1348 Problem Description Connecting the display screen and signal sources which produce differen

hdu6195 cable cable cable(from 2017 ACM/ICPC Asia Regional Shenyang Online)

最开始一直想不通,为什么推出这个公式,后来想了半天,终于想明白了. 题目大意是,有M个格子,有K个物品.我们希望在格子与物品之间连数量尽可能少的边,使得——不论是选出M个格子中的哪K个,都可以与K个物品恰好一一匹配. 然后你可以试着画图,每次必须有k个格子是单独的(与各物体只有一条线相连)所以还剩下m-k个格子,可以用来补位,也就是跟每个物品都相连,所以就有(m-k)*k 上代码(巨巨巨巨巨简单): 1 #include <cstdio> 2 #include <cstring>

2017 ACM/ICPC Asia Regional Shenyang Online 记录

这场比赛全程心态爆炸…… 开场脑子秀逗签到题WA了一发.之后0贡献. 前期状态全无 H题想复杂了,写了好久样例过不去. 然后这题还是队友过的…… 后期心态炸裂,A题后缀数组理解不深,无法特判k = 1时的情况. 然后也没有心思读题了,心静不下来. 比赛题目链接 Problem B $ans = k(n - k + 1)$ #include <bits/stdc++.h> using namespace std; typedef long long LL; LL n, k; int main()

2017 ACM/ICPC Asia Regional Qingdao Online 1011 A Cubic number and A Cubic Number

A Cubic number and A Cubic Number Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description A cubic number is the result of using a whole number in a mul