待通过：A、D、G、J、M
已补过：E
E:电路题，判断一个图是不是简单电路。不需要特殊的技巧，利用set存图，把度数为2的点都删掉，融入到一条边上即可。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 9999973; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 1e5+9; set<int>mp[maxn]; queue<int>que; int vis[maxn]; int main(){ int n,m; scanf("%d%d", &n, &m); for(int i=1; i<=m; i++){ int u,v; scanf("%d%d", &u, &v); mp[u].insert(v); mp[v].insert(u); } for(int i=1; i<=n; i++){ if(mp[i].size() == 2) que.push(i); } while(!que.empty()){ int u = que.front(); que.pop(); if(mp[u].size() != 2) continue; int s = *mp[u].begin(); mp[u].erase(s); int x = *mp[u].begin(); mp[u].erase(x); mp[s].erase(u); mp[x].erase(u); mp[s].insert(x); mp[x].insert(s); if(mp[s].size() == 2) que.push(s); if(mp[x].size() == 2) que.push(x); } int flag = 1,cnt = 0; for(int i=1; i<=n; i++){ if(mp[i].size() >= 2) flag = 0; else if(mp[i].size() == 1) cnt ++; } //debug(cnt); if(flag && cnt == 2) puts("Yes"); else puts("No"); return 0; }