Spanning Tree with One Fixed Degree

https://codeforces.com/contest/1133/problem/F2

题解：先把与1无关的点合并，然后找与一相关必须连接的边，对于答案树想连必须的边，再连其他一相关边，最后连其他边

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp,sum,u,v,d;
int a[N];
int pre[N];
int preans[N];
int vis[N];
bool broken[N];
char str;
vector<int>G[N];
struct Node{int u,v;bool operator <(const Node&S)const{if(u==S.u)return v<S.v;return u<S.u;}
}x;
vector <Node>ANS;
int find(int x){return (pre[x]==x)?x:pre[x]=find(pre[x]);}
int findans(int x){return (preans[x]==x)?x:preans[x]=findans(preans[x]);}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d%d",&n,&m,&d);for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);G[u].push_back(v);G[v].push_back(u);}if((int)G[1].size()<d){cout<<"NO"<<endl;return 0;}for(int i=1;i<=n;i++){pre[i]=i;preans[i]=i;}for(int i=2;i<=n;i++){for(int j=0,k=G[i].size();j<k;j++){u=i;v=G[i][j];if(v==1)continue;int tx=find(u);int ty=find(v);if(tx!=ty){pre[tx]=ty;}}}for(int i=0,j=G[1].size();i<j;i++){u=1;v=G[1][i];int tx=find(u);int ty=find(v);if(tx!=ty){pre[tx]=ty;//cout<<u<<" "<<v<<endl;x.u=u;x.v=v;int ta=findans(u);int tb=findans(v);preans[ta]=tb;ANS.push_back(x);if(++cnt>=d)break;}}for(int i=0,j=G[1].size();i<j;i++){u=1;v=G[1][i];int ta=findans(u);int tb=findans(v);if(cnt>=d)break;if(ta!=tb){//cout<<u<<" "<<v<<endl;cnt++;x.u=u;x.v=v;preans[ta]=tb;ANS.push_back(x);}}for(int i=2;i<=n;i++){for(int j=0,k=G[i].size();j<k;j++){u=i;v=G[i][j];if(v==1)continue;int ta=findans(u);int tb=findans(v);if(ta!=tb){//cout<<u<<" "<<v<<endl;x.u=u;x.v=v;preans[ta]=tb;ANS.push_back(x);}}}flag=0;for(int i=1;i<=n;i++){if(pre[i]==i){flag++;}}if(flag>1){cout<<"NO"<<endl;}else{cout<<"YES"<<endl;sort(ANS.begin(),ANS.end());for(int i=0,j=ANS.size();i<j;i++){x=ANS[i];cout<<x.u<<" "<<x.v<<endl;}}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

C++版本二

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAX_N=2e5+10;
int par[MAX_N];     //父亲
int ranks[MAX_N];    //树的高度
struct Edge{int u,v;};
Edge edges[MAX_N];
int cnt[MAX_N]={0,};
int n,m,k;
int ans[MAX_N],q=0;
void init(){for(int i=1;i<=n;i++){par[i]=i;ranks[i]=0;}
}int findit(int x){if(par[x]==x)return x;elsereturn par[x]=findit(par[x]);
}void unite(int x, int y){x=findit(x);y=findit(y);if(x==y) return;if(ranks[x]<ranks[y])par[x]=y;else par[y]=x;if(ranks[x]==ranks[y])ranks[x]++;
}bool same(int x, int y){return findit(x)==findit(y);
}int main(void){scanf("%d %d %d",&n,&m,&k);init();for(int i=0;i<m;++i){scanf("%d %d",&edges[i].u,&edges[i].v);cnt[edges[i].u]++;cnt[edges[i].v]++;}if(cnt[1]<k){printf("NO\n");return 0;}//先把剩下的点连起来for(int i=0;i<m;++i)if(edges[i].u!=1&&edges[i].v!=1)if(!same(edges[i].u,edges[i].v))unite(edges[i].u,edges[i].v);//得到顶点1必须选用的边for(int i=0;i<m;++i){if(edges[i].u==1||edges[i].v==1){if(!same(edges[i].u,edges[i].v)){unite(edges[i].u,edges[i].v);ans[q++]=i;}}}if(q>k){printf("NO\n");return 0;}//重新初始化init();printf("YES\n");for(int i=0;i<q;++i){unite(edges[ans[i]].u,edges[ans[i]].v);printf("%d %d\n",edges[ans[i]].u,edges[ans[i]].v);}for(int i=0;i<m&&q<k;++i){if(edges[i].u==1||edges[i].v==1){if(!same(edges[i].u,edges[i].v)){unite(edges[i].u,edges[i].v);printf("%d %d\n",edges[i].u,edges[i].v);q++;}}}//再连其他点for(int i=0;i<m;++i){if(edges[i].u!=1&&edges[i].v!=1){if(!same(edges[i].u,edges[i].v)){unite(edges[i].u,edges[i].v);printf("%d %d\n",edges[i].u,edges[i].v);}}}
}

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