大意:$n$結點,$m$條邊無向圖, 有$k$個人, 每個人最多走$\left\lceil\frac {2n}{k}\right\rceil$步, 求一種方案使得$k$個人走遍所有的點

$n$結點樹的歐拉序長度為$2n-1$, 直接取$dfs$樹的歐拉序即可

#include <iostream>
#include <algorithm>
#include <math.h>
#include <cstdio>
#include <vector>
#define pb push_back
#define REP(i,a,n) for(int i=a;i<=n;++i)
using
 
 namespace std;

const int N = 4e5+10, INF = 0x3f3f3f3f;
int n, m, k, mx;
int vis[N];
vector<int> g[N], path;

void dfs(int x) {
    vis[x] = 1, path.pb(x);
    for (int y:g[x]) {
        if (!vis[y]) dfs(y),path.pb(x);
    }
}

int main() {
    cin>>n>>m>>k;
    REP(i,1,m) {
        
 
int x, y;
        cin>>x>>y;
        g[x].pb(y),g[y].pb(x);
    }
    dfs(1),path.pop_back();
    for (int i=1; i<=k; cout<<‘\n‘,++i) {
        if (path.empty()) { 
            cout<<"1 1";
            continue;
        }
        int mx = (2*n+k-1)/k;
        if (mx>path.size()) mx = path.size();
        cout
 
<<mx<<‘ ‘;
        REP(j,1,mx) {
            cout<<path.back()<<‘ ‘;
            path.pop_back();
        }
    }
}

Underground Lab CodeForces - 782E (歐拉序)