cf 【并查集】】的更多相关文章

CF A. Party time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who…
题目地址:24道CF的DIv2 CD题有兴趣可以做一下. ACM思维题训练集合 Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He ha…
题目:http://codeforces.com/problemset/problem/437/D 题意:有n个点,m条边的无向图,保证所有点都能互通,n,m<=10^5 每个点都有权值,每条边的权值定义为这条边连接两点的权值中的最小值. f(p,q)表示p到q的路径中边权的最小值,如果有多条路经,就取每条路径最小值中的最小值 要求的就是对于所有1<=p<=n,1<=q<=n,p≠q,f(p,q)的平均值 分析: 刚开始是想考虑每条边对总和的贡献,结果没想通. 一个很巧的办法…
User ainta has a permutation p1, p2, ..., pn. As the New Year is coming, he wants to make his permutation as pretty as possible. Permutation a1, a2, ..., an is prettier than permutation b1, b2, ..., bn, if and only if there exists an integer k (1 ≤ k…
C. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a re…
C. Glass Carving time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a re…
https://codeforces.com/contest/1257/problem/E 题意:有三个集合集合里面的数字可以随意变换位置,不同集合的数字,如从第一个A集合取一个数字到B集合那操作数+1,求最少操作次数使得一二三集合连起来是升序 我们先把每个集合排个序,然后拼起来,求一遍lis(最长升序序列:https://blog.csdn.net/Joined/article/details/78058362),然后n-len即为答案 网上那些什么dp什么的..嘻嘻太高深 #include<…
http://codeforces.com/contest/1245/problem/D 题意就是:你需要让所有城市都有电,你看也在该城市建电站使他有电,同时你可以链接他与其他城市,使之有电 解决: 我们可以吧每个城市自己建电站以及自己与其他城市的费用用结构体存起来,排个序,再用并查集连起来. #include<bits/stdc++.h> #define numm ch-48 #define pd putchar(' ') #define pn putchar('\n') #define p…
$des$ 题面 $sol$ 把边从小到大排序,枚举每条边作为答案,然后把两个点合并,判断每条边是否可以作为答案时,$cnt_i$ 表示节点 $i$ 已经合并的 $x$ 之和$size_i$ 表示已经合并的节点的个数$sum = \sum x$将 $a$ 与外面的点合并时判断条件 $size_a <= sum - cnt_a$并查集维护. $code$ #include <bits/stdc++.h> using namespace std; #define Rep(i, a, b) f…
传送门 解题思路 感觉这种题都是套路之类的??首先把三个串并成一个,中间插入一些奇怪的字符,然后跑遍\(SA\).考虑按照\(height\)分组计算,就是每个\(height\)只在最高位计算一次,然后求个后缀和,这个可以并查集来实现.具体就是记一个\(sum[i][3]\)表示第\(i\)个联通块中\(0,1,2\)的个数,\(0,1,2\)就是出现在三个串的哪一个,然后合并时需要容斥一下. 代码 #include<iostream> #include<cstdio> #inc…