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_queuetypedef 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; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = ;
const double esp = 1e-;
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=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
}
/*-----------------------showtime----------------------*/ const int maxn = 1e5+;
set<int>mp[maxn];
queue<int>que;
int vis[maxn];
int main(){
int n,m;
scanf("%d%d", &n, &m);
for(int i=; i<=m; i++){
int u,v;
scanf("%d%d", &u, &v);
mp[u].insert(v);
mp[v].insert(u);
} for(int i=; i<=n; i++){
if(mp[i].size() == ) que.push(i);
}
while(!que.empty()){ int u = que.front(); que.pop();
if(mp[u].size() != ) 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() == ) que.push(s);
if(mp[x].size() == ) que.push(x);
}
int flag = ,cnt = ;
for(int i=; i<=n; i++){
if(mp[i].size() >= ) flag = ;
else if(mp[i].size() == ) cnt ++;
}
//debug(cnt);
if(flag && cnt == ) puts("Yes");
else puts("No");
return ;
}
E
