Codeforces Round #121 (Div. 1) C. Fools and Roads time limit per test :2 seconds memory limit per test : 256 megabytes They say that Berland has exactly two problems, fools and roads. Besides, Berland has n cities, populated by the fools and connecte…

题目链接:Codeforces 191C Fools and Roads 题目大意:给定一个N节点的数.然后有M次操作,每次从u移动到v.问说每条边被移动过的次数. 解题思路:树链剖分维护边,用一个数组标记就可以,不须要用线段树. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 1e5 + 5; int N, Q, ne, fi…

主题链接~~> 做题情绪:做了HDU 5044后就感觉非常easy了. 解题思路: 先树链剖分一下,把树剖分成链,由于最后全是询问,so~能够线性操作.经过树链剖分后,就会形成很多链,可是每条边都有编号,相当于一个数组进行线性操作,这样.如果在 u  ~ v 去都添加 1 .那么能够让 sum [ u ] += 1 ; sum [ v + 1 ] -= 1 ; 这里如果 v 的编号大. 最后的时候仅仅要从后往前遍历一次就能够了.得到全部的结果.明确这点后再加上树链剖分的思想就能够攻克了. 代码:…

题目大意:有一颗$n$个节点的树,$k$次旅行,问每一条被走过的次数. 题解:树上差分,$num_x$表示连接$x$和$fa_x$的边被走过的次数,一条路径$u->v$,$num_u+1,num_v+1,num_{LCA(u,v)}-2$,最后求个子树和就行了 卡点:无 C++ Code: #include <cstdio> #include <algorithm> #define maxn 100010 int head[maxn], cnt = 1; struct Edg…

They say that Berland has exactly two problems, fools and roads. Besides, Berland has n cities, populated by the fools and connected by the roads. All Berland roads are bidirectional. As there are many fools in Berland, between each pair of cities th…

一道经典的带修改树链第 \(k\) 大的问题. 我只想出三个 \(\log\) 的解法... 整体二分+树剖+树状数组. 那不是暴力随便踩的吗??? 不过跑得挺快的. \(Code\ Below:\) // luogu-judger-enable-o2 #include <bits/stdc++.h> #define lowbit(x) ((x)&(-(x))) using namespace std; const int maxn=80000+10; const int lim=1e…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6162 题意:给出一棵树的链接方法,每个点都有一个数字,询问U->V节点经过所有路径中l < = x < = r的数字和 解法:主席树维护区间和,树剖查询,复杂度nloglog. 代码: #include <bits/stdc++.h> using namespace std; const int maxn = 1e5+5; const int maxm = 40*maxn; ty…

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…

题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1307 题意: 中文题诶~ 思路: 解法1:暴力树剖 用一个数组 num[i] 维护编号为 i 的边当前最大能承受的重量. 在加边的过程中根据给出的父亲节点将当前边所在的链上所有边的num都减去当前加的边的重量, 注意当前边也要减自重. 那么当num首次出现负数时加的边号即位答案: 事实上这个算法的时间复杂度是O(n^2)的, 不过本题并没有出那种退化成单链的…

After their royal wedding, Jiajia and Wind hid away in XX Village, to enjoy their ordinary happy life. People in XX Village lived in beautiful huts. There are some pairs of huts connected by bidirectional roads. We say that huts in the same pair dire…