gym/102059 E

gym/102059

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;
}

E

原文地址：https://www.cnblogs.com/ckxkexing/p/10324234.html