题目链接：https://vjudge.net/problem/CodeForces-1176E
题目描述：
You are given an undirected unweighted connected graph consisting of nn vertices and mm edges. It is guaranteed that there are no self-loops or multiple edges in the given graph.
Your task is to choose at most ⌊n2⌋⌊n2⌋ vertices in this graph so each unchosen vertex is adjacent (in other words, connected by an edge) to at least one of chosen vertices.
It is guaranteed that the answer exists. If there are multiple answers, you can print any.
You will be given multiple independent queries to answer.
Input
The first line contains a single integer tt (1≤t≤2⋅1051≤t≤2⋅105) — the number of queries.
Then tt queries follow.
The first line of each query contains two integers nn and mm (2≤n≤2⋅1052≤n≤2⋅105, n−1≤m≤min(2⋅105,n(n−1)2)n−1≤m≤min(2⋅105,n(n−1)2)) — the number of vertices and the number of edges, respectively.
The following mm lines denote edges: edge ii is represented by a pair of integers vivi, uiui (1≤vi,ui≤n1≤vi,ui≤n, ui≠viui≠vi), which are the indices of vertices connected by the edge.
There are no self-loops or multiple edges in the given graph, i. e. for each pair (vi,uivi,ui) there are no other pairs (vi,uivi,ui) or (ui,viui,vi) in the list of edges, and for each pair (vi,uivi,ui) the condition vi≠uivi≠ui is satisfied. It is guaranteed that the given graph is connected.
It is guaranteed that ∑m≤2⋅105∑m≤2⋅105 over all queries.
Output
For each query print two lines.
In the first line print kk (1≤⌊n2⌋1≤⌊n2⌋) — the number of chosen vertices.
In the second line print kk distinct integers c1,c2,…,ckc1,c2,…,ck in any order, where cici is the index of the ii-th chosen vertex.
It is guaranteed that the answer exists. If there are multiple answers, you can print any.
Example
2 4 6 1 2 1 3 1 4 2 3 2 4 3 4 6 8 2 5 5 4 4 3 4 1 1 3 2 3 2 6 5 6
2 1 3 3 4 3 6
Note
In the first query any vertex or any pair of vertices will suffice.
Note that you don't have to minimize the number of chosen vertices. In the second query two vertices can be enough (vertices 22 and 44) but three is also ok.
1.通过判断子节点与父节点来选取数量较少的一边： #include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl '\n' #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define zero(a) memset(a, 0, sizeof(a)) #define INF(a) memset(a, 0x3f, sizeof(a)) #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; typedef pair<double, int> P2; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1000000007LL; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const double EULC = 0.5772156649015328; const int NIL = -1; template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } const int maxn = 2e5+10; int vis[maxn]; vector<int> eage[maxn], odd, even; void dfs(int u, int flag) { if (flag && !vis[u]) odd.push_back(u); else even.push_back(u); vis[u] = true; for (auto node : eage[u]) if (!vis[node]) dfs(node, !flag); } int main(void) { int t; scanf("%d", &t); while(t--) { int n, m; scanf("%d%d", &n, &m); for (int i = 0, u, v; i<m; ++i) { scanf("%d%d", &u, &v); eage[u].push_back(v); eage[v].push_back(u); dfs(1, 1); int size1 = odd.size(), size2 = even.size(); printf("%d\n", min(size1, size2)); if (size1 < size2) for (int i = 0; i<size1; ++i) printf(i == size1-1 ? "%d\n" : "%d ", odd[i]); else for (int i = 0; i<size2; ++i) printf(i == size2-1 ? "%d\n" : "%d ", even[i]); for (int i = 0; i<=n; ++i) eage[i].clear(), vis[i] = 0; odd.clear(), even.clear(); } return 0; 2.通过判断与某个结点距离的奇偶性 #include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl '\n' #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define zero(a) memset(a, 0, sizeof(a)) #define INF(a) memset(a, 0x3f, sizeof(a)) #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<ll, ll> P; typedef pair<ll, ll> P2; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1000000007LL; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const double EULC = 0.5772156649015328; const int NIL = -1; template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } const int maxn = 2e5+10; int d[maxn]; vector<int> odd, even, eage[maxn]; void bfs() { queue<int> qe; d[1] = 0; qe.push(1); even.push_back(1); while(!qe.empty()) { int t = qe.front(); qe.pop(); for (auto to : eage[t]) if (d[to] == INF) { d[to] = d[t]+1; qe.push(to); if (d[to]&1) odd.push_back(to); else even.push_back(to); } } } int main(void) { int t; scanf("%d", &t); while(t--) { int n, m; scanf("%d%d", &n, &m); for (int i = 0; i<n+10; ++i) d[i] = INF; for (int i = 0, u, v; i<m; ++i) { scanf("%d%d", &u, &v); eage[u].push_back(v); eage[v].push_back(u); } bfs(); int size1 = odd.size(), size2 = even.size(); printf("%d\n", min(size1, size2)); if (size1 < size2) for (int i = 0; i<size1; ++i) printf(i == size1-1 ? "%d\n" : "%d ", odd[i]); else for (int i = 0; i<size2; ++i) printf(i == size2-1 ? "%d\n" : "%d ", even[i]); for (int i = 0; i<=n; ++i) eage[i].clear(); odd.clear(), even.clear(); } return 0; }