Codeforces Round #403 (Div. 2) E. Underground Lab

题意:给n点m边的连通图，k个人从任意点出发至多经过  个点,求分配方式使得每个点都至少被走过一次

题解：在图中以任意点为根任意dfs出一个生成树，每条边走2次（也就是每次向前走完再回头走），按dfs的顺序得出一个答案序列，总共走过2*(n-1)+1 = 2*n-1个点并且走完所有点，再把连续的  个点分配给一个人，多的人输出1 1即可

#include <iostream>#include <cstdio>#include <cctype>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <set>#include <stack>#include <sstream>#include <queue>#include <map>#include <functional>#include <bitset>using namespace std;#define pb push_back#define ll long long#define ull unsigned long long#define pii pair<int, int>#define mk make_pair#define fi first#define se second#define ALL(A) A.begin(), A.end()#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)#define repr(i, n) for(int (i)=(int)(n);(i)>=0;(i)--)#define repab(i,a,b) for(int (i)=(int)(a);(i)<=(int)(b);(i)++)#define reprab(i,a,b) for(int (i)=(int)(a);(i)>=(int)(b);(i)--)#define sc(x) scanf("%d", &x)#define pr(x) printf("x:%d\n", x)#define fastio ios::sync_with_stdio(0), cin.tie(0)#define frein freopen("in.txt", "r", stdin)#define freout freopen("out.txt", "w", stdout)#define freout1 freopen("out1.txt", "w", stdout)#define lb puts("")#define PI M_PI#define debug cout<<"???"<<endl#define mid ((l+r)>>1)const ll mod = 1000000007;//const int INF = 0x3f3f3f3f;const ll INF = 0x3f3f3f3f3f3f3f3f;const double eps = 1e-6;template<class T> T gcd(T a, T b){if(!b)return a;return gcd(b,a%b);}const int maxn = 2e5+10;vector<int> ans, edg[maxn];int n,m,k,vis[maxn];void dfs(int u){    vis[u] = 1;    ans.pb(u);    for(int i = 0; i < edg[u].size(); i++){        int v = edg[u][i];        if(vis[v]) continue;        dfs(v);        ans.pb(u);    }}int main(){    //frein;    cin >> n >> m >> k;    for(int i = 0; i < m; i++){        int u,v; sc(u); sc(v);        edg[u].pb(v); edg[v].pb(u);    }    dfs(1);    int stp = (2*n+k-1)/k;    for(int i = 0; i < k; i++){        int beg = i*stp, ed = min((i+1)*stp, (int)ans.size());        if(ed <= beg){            printf("1 1\n");            continue;        }        printf("%d", ed-beg);        for(int j = beg; j < ed; j++){            printf(" %d", ans[j]);        }        lb;    }    return 0;}