nxt

不能说是一个算法，应该算是一类思想 点分治 概念 点分治就是把树上问题中的节点拿来 分治 。 这所谓的“分治”是一个很抽象的概念，那么就先来介绍它的常见应用和其他性质。 大致框架 1 void getRoot(int x, int fa) //找重心 2 { 3 size[x] = 1, son[x] = 0; 4 for (int i=head[x]; i!=-1; i=nxt[i]) 5 { 6 int v = edges[i].y; 7 if (v==fa||vis[v]) continue; //在点分树内寻找重心 8 getRoot(v, x), size[x] += size[v]; 9 son[x] = std::max(son[x], size[v]); 10 } 11 son[x] = std::max(son[x], tot-size[x]); 12 if (son[x] < son[root]) root = x; //root即重心 13 } 14 void dfs(int x, int fa, int c) 15 { 16 record distance_c //将长度为c的路径记录下来 17 for (int i=head[x]; i!=-1; i=nxt[i]) 18 { 19 int v = edges[i].y; 20 if (v==fa||vis

题目链接： http://poj.org/problem?id=1741 Time Limit: 1000MS Memory Limit: 30000K Description Give a tree with n vertices, each edge has a length(positive integer less than 1001). Define dist(u,v)=The min distance between node u and v. Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k. Write a program that will count how many pairs which are valid for a given tree. Input The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which

7-52 两个有序链表序列的交集 (20分) AC代码 # include <iostream> # include <cstdio> # include <cstdlib> using namespace std ; typedef struct node { int data ; struct node * nxt ; } node , * linklist ; linklist Read ( ) { linklist L , op ; L = ( linklist ) malloc ( sizeof ( node ) ) ; L -> data = - 1 ; L -> nxt = NULL ; op = L ; int num ; while ( scanf ( "%d" , & num ) && num != - 1 ) { linklist p = ( linklist ) malloc ( sizeof ( node ) ) ; p -> data = num ; p -> nxt = NULL ; op -> nxt = p ; op = op -> nxt ; } op -> nxt = NULL ; return L ; } linklist Common ( linklist la , linklist lb ) { linklist L , op , pa

7-51 两个有序链表序列的合并 (20分) AC代码 # include <iostream> # include <cstdio> # include <vector> # include <algorithm> # include <cstdlib> using namespace std ; typedef struct node { int data ; struct node * nxt ; } node , * linklist ; linklist Read ( ) { linklist L , op ; L = ( linklist ) malloc ( sizeof ( node ) ) ; L -> data = - 1 ; L -> nxt = NULL ; op = L ; int data ; scanf ( "%d" , & data ) ; while ( data != - 1 ) { linklist tmp = ( linklist ) malloc ( sizeof ( node ) ) ; tmp -> nxt = NULL ; tmp -> data = data ; op -> nxt = tmp ; op = op -> nxt ; scanf ( "%d" , & data ) ; } return L ; } linklist

好，暴力能拿$50pts\space qwq$ 暴力的思路就是一直跳$nxt[j]$，直到它的长度小于串的一半，然后开始计数，当然要接着跳$nxt[j]$ 正解：考虑没有长度要求的（不要求不重合）公共前后缀的数目，显然$ans[i]=ans[j]+1$相当于$i$比$j$是多了$i$它本身。 所以求解时模仿$kmp$的过程，$num[i]$就是一直跳$nxt[j]$,然后直到$j<=\frac{i}{2}$,$num[i]=ans[j]$ 注意，此处模仿$kmp$的原因是，能够避免跳过一些冗余的$nxt[j]$（$nxt[j]$过大的情况）,对于下一次匹配，上一次的$j$就是上一次的小于等于$\frac{i}{2}$的最长公共前后缀的长度，如果还能匹配就直接匹配，过长了再跳一下$nxt[j]$，否则就一直跳$nxt[j]$，直到匹配。 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cctype> #include<cstdlib> #include<vector> #include<queue> #include<map> #include<set> #define ull unsigned long long #define ll

# 117. 填充每个节点的下一个右侧节点指针 II # https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii from copy import copy class Solution: def connect(self, root: 'Node') -> 'Node': if root is None: return root cur, nxt = [], [] cur.append(root) root.next = None while cur: while cur: tmp = cur.pop(0) if tmp.left: nxt.append(tmp.left) if tmp.right: nxt.append(tmp.right) if len(nxt) != 0: for i in range(len(nxt) - 1): nxt[i].next = nxt[i + 1] nxt[len(nxt) - 1].next = None # 这里如果用deepcopy的话，你会爽歪歪！！！ 你要用原来的树节点，而不是deepcopy创建的新节点 cur = copy(nxt) nxt.clear() return root 来源： CSDN 作者： 踩着七彩祥云的猴子