这道题有一种较为暴力的做法,对于每个点枚举所有与r2为该属性的询问并加以修改,最坏时间复杂度为o(nq),然而是可过的(97s) 发现只有当r2相同的询问数特别多时才会达到最坏时间复杂度,因此如果删除重复询问,时间复杂度降为o(nr),然而并没有显著优化(81s) 接着考虑当同一种r2的询问特别多时(大于K),可以从r1考虑,分析一下时间复杂度发现是$o(n\cdot max(K,R/K))\ge o(n\sqrt{R})$,即取$K=\sqrt{R}$,此时只要13s 1 #include<b…

思路:首先如果颜色相同直接利用以前的答案即可,可以离线排序或是在线hash,然后考虑怎么快速统计答案. 首先如果点a是点b的祖先,那么一定有点b在以点a为根的子树的dfs序区间内的,于是先搞出dfs序. 然后如果颜色a的点数很小,颜色b的点数很大,那么可以考虑枚举a的点数,然后对于每一种颜色开个vector记录一下有哪些点是这种颜色,然后按照它们的dfs序排序,就可以用颜色a中的每个点在颜色b中二分出哪些点属于以该点为根的子树对应的dfs序区间了.复杂度O(size(a)*log(size(b)…

题意 题目链接 Sol 很神仙的题 我们考虑询问(a, b)(a是b的祖先),直接对b根号分治 如果b的出现次数\(< \sqrt{n}\),我们可以直接对每个b记录下与它有关的询问,这样每个询问至多扫\(\sqrt{n}\)个点即可知道答案,那么dfs的时候暴力统计答案即可,复杂度\(q\sqrt{n}\) 如果b的出现次数\(> \sqrt{n}\),显然这样的b最多只有\(\sqrt{n}\)个,也就是说在询问中最多会有\(\sqrt{n}\)个这样的b,那么我们可以对每个a,暴力统计,…

http://dzy493941464.sinaapp.com/archives/96 那个SIZE貌似必须设成R*R/Q?不知为啥,自己算的不是这个的说. 本机AC,线上TLE. #include<cstdio> #include<set> #include<vector> #include<cmath> #include<algorithm> using namespace std; int f,C; void R(int &x){…

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region. For example, X X X X X O O X X X O X X O X X After running your function, the board should…

在LeetCode中的Surrounded Regions 包围区域这道题中,我们发现用DFS方法中的最后一个条件必须是j > 1,如下面的红色字体所示,如果写成j > 0的话无法通过OJ,一直百思不得其解其中的原因,直到有网友告诉我说他验证了最后一个大集合在本地机子上可以通过,那么我也来验证看看吧. class Solution { public: void solve(vector<vector<char> >& board) { ; i < boar…

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region . For example, X X X X X O O X X X O X X O X X After running your function, the board should…

Total Accepted: 43584 Total Submissions: 284350 Difficulty: Medium Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region. For example, X X X X X O…

When it comes to Amazon Web Services, there are two concepts that are extremely important and spanning across all the services, and that you simply can’t help but be aware of: Regions and Availability Zones. Both of them associate with most of the AW…

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'. A region is captured by flipping all 'O's into 'X's in that surrounded region. For example, X X X X X O O X X X O X X O X X After running your function, th…