dfs判断图的连通块数量~ #include<cstdio> #include<algorithm> #include<vector> #include<cstring> using namespace std; ; vector<int> g[maxn]; int visit[maxn],N,M,x,y,flag; void init () { fill (visit,visit+maxn,); ;i<maxn;i++) g[i].cle…

E. Cyclic Components time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an undirected graph consisting of nn vertices and mm edges. Your task is to find the number of connected…

Cyclic Components CodeForces - 977E You are given an undirected graph consisting of nn vertices and mm edges. Your task is to find the number of connected components which are cycles. Here are some definitions of graph theory. An undirected graph con…

E. Cyclic Components time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an undirected graph consisting of nn vertices and mm edges. Your task is to find the number of connected…

You are given an undirected graph consisting of nn vertices and mm edges. Your task is to find the number of connected components which are cycles. Here are some definitions of graph theory. An undirected graph consists of two sets: set of nodes (cal…

题意 给出nnn个顶点和mmm条边,求这个图中环的个数 思路 利用并查集的性质,环上的顶点都在同一个集合中 在输入的时候记录下来每个顶点的度数,查找两个点相连,且度数均为222的点,如果这两个点的父节点相同,表示这两个点在一个环中,环的个数+1+1+1 AC代码 /************************************************************************* > File Name: E.cpp > Author: WZY > QQ:…

题意: 就是找出所有环的个数, 但这个环中的每个点都必须只在一个环中 解析: 在找环的过程中 判断度数是否为2就行...emm... #include <bits/stdc++.h> using namespace std; , INF = 0x7fffffff; int n, m, s, t, flag; vector<int> G[maxn]; int vis[maxn], res; void dfs(int u, int fa) { vis[u] = ; ; i<G[u…

Description You are given an undirected graph consisting of nn vertices and mm edges. Your task is to find the number of connected components which are cycles.Here are some definitions of graph theory.An undirected graph consists of two sets: set of…

#include <bits/stdc++.h> using namespace std; *1e5+; vector<int>p[maxn]; vector<int>c; bool check[maxn]; void dfs(int x) { check[x] = true; c.push_back(x); ; i < p[x].size(); i++) { int y = p[x][i]; if (!check[y]) { dfs(y); } } } int…

题意:给你\(n\)个顶点和\(m\)条边,问它们有多少个单环(无杂环),例如图中第二个就是一个杂环. 题解:不难发现,如果某几个点能够构成单环,那么每个点一定只能连两条边.所以我们先构建邻接表,然后从某个数开始跑dfs,如果这一边所有点的度数都为\(2\),那么就能构成一个单环. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include &l…