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

題解：先把與1無關(guān)的點合并，然后找與一相關(guān)必須連接的邊，對于答案樹想連必須的邊，再連其他一相關(guān)邊，最后連其他邊

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]; //父親 int ranks[MAX_N]; //樹的高度 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;}//先把剩下的點連起來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);//得到頂點1必須選用的邊f(xié)or(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++;}}}//再連其他點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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