题目大意:给定一个 N 个点的有向图,计数图上哈密顿回路的条数. 题解:哈密顿回路需要经过除了初始位置,每个点恰好一次.如果已知一条哈密顿回路的方向,那么从这条路上任意一个点出发,得到的都是同样的结果.因此,不妨设从 0 号节点出发,最后回到 0 号节点.统计答案只需要枚举最后一个点在哪个位置即可. 代码如下 #include <bits/stdc++.h> using namespace std; typedef long long LL; int n,m,G[12][12]; LL f[1…

原题链接:CF1205B   题目大意   给定\(n\)个整数\(a_1,a_2,a_3, \dots ,a_n\),若\(i \neq j\)且\(a_i \land a_j \neq 0\),则节点\(i\)和节点\(j\)相连通.求最小环大小.   \(1 \leq n \leq 10^5\),\(0 \leq a_i \leq 10^{18}\).   题解   题目中的\(a_i\)都在long long的范围内,即\(a_i \in [0, 2^{64})\),则根据容斥原理可得,…

The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle". In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle. Input Specif…

Difficulty:medium  More:[目录]LeetCode Java实现 Description Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? Intuiti…

142. Linked List Cycle II[easy] Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 解法一: /** * Definition for singl…

141. Linked List Cycle[easy] Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x…

[题解]cycle 题目描述 给定一个无向图,求一个环,使得环内边权\(\div\)环内点数最大. 数据范围 \(n \le 5000\) \(m\le 10000\) \(Solution\) 考虑到我们可以对答案的式子变一下形, \(\frac{\Sigma_{i\in V'} w_i}{|V'|}\le ans\) \(\Sigma_{i\in V'}w_i-ans\times |V'|\le0\) 这一步不要看不懂了(\(i\)共有\(|V'|\)个,所以\(ans\)仍然总共被加了\(…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 思路:由[Leetcode]Linked List Cycle可知.利用一快一慢两个指针可以推断出链表是否存在环路. 如果两个指针相遇之前slow走了s步,则fast走了2s步.而且fast已经在长…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] p[i] = p[p[i]]一直进行下去 在1..n的排列下肯定会回到原位置的. 即最后会形成若干个环. g[i]显然等于那个环的大小. 即让你形成若干个环. 每个环的大小只能为A或B 则相当于问Ax+By=n是否有解. 可以枚举x然后看看n-A*x能否被B整除. 构造x个长度为A的环,y个长度为B的环就好了 [代码] #include <bits/stdc++.h> using namespace std; const in…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output A group of n cities is connected by a network of roads. There is an undirected road between every pair of cities, so there are roads in total. I…