【Codeforces Round #403】Codeforces 781C Underground Lab

The evil Bumbershoot corporation produces clones for gruesome
experiments in a vast underground lab. On one occasion, the corp
cloned a boy Andryusha who was smarter than his comrades. Immediately
Andryusha understood that something fishy was going on there. He
rallied fellow clones to go on a feud against the evil corp, and they
set out to find an exit from the lab. The corp had to reduce to
destroy the lab complex.

The lab can be pictured as a connected graph with n vertices and m
edges. k clones of Andryusha start looking for an exit in some of the
vertices. Each clone can traverse any edge once per second. Any number
of clones are allowed to be at any vertex simultaneously. Each clone
is allowed to stop looking at any time moment, but he must look at his
starting vertex at least. The exit can be located at any vertex of the
lab, hence each vertex must be visited by at least one clone.

Each clone can visit at most vertices before the lab explodes.

Your task is to choose starting vertices and searching routes for the
clones. Each route can have at most vertices. Input

The first line contains three integers n, m, and k (1 ≤ n ≤ 2·105,
n - 1 ≤ m ≤ 2·105, 1 ≤ k ≤ n) — the number of vertices and edges in
the lab, and the number of clones.

Each of the next m lines contains two integers xi and yi
(1 ≤ xi, yi ≤ n) — indices of vertices connected by the respective
edge. The graph is allowed to have self-loops and multiple edges.

The graph is guaranteed to be connected. Output

You should print k lines. i-th of these lines must start with an
integer ci () — the number of vertices visited by i-th clone, followed
by ci integers — indices of vertices visited by this clone in the
order of visiting. You have to print each vertex every time it is
visited, regardless if it was visited earlier or not.

It is guaranteed that a valid answer exists.

如果只有一个人怎么走？这不就是欧拉序，经过点数是2n−1。
有k个人呢？你把一个人的路径拆开不就行了。

#include<cstdio>#include<algorithm>using namespace std;int fir[200010],ne[400010],to[400010],vis[200010],f[400010],num[200010],n,m,k,clo;void add(int num,int u,int v){    ne[num]=fir[u];    fir[u]=num;    to[num]=v;}void dfs(int u){    int v;    f[++clo]=u;    vis[u]=1;    for (int i=fir[u];i;i=ne[i])        if (!vis[v=to[i]])        {            dfs(v);            f[++clo]=u;        }}int main(){    int x,y;    scanf("%d%d%d",&n,&m,&k);    for (int i=1;i<=m;i++)    {        scanf("%d%d",&x,&y);        add(i*2,x,y);        add(i*2+1,y,x);    }    dfs(1);    for (int i=1;i<=k;i++) num[i]=(2*n-1)/k;    for (int i=1;i<=(2*n-1)%k;i++) num[i]++;    for (int i=1,j=1;i<=k;i++)    {        printf("%d ",num[i]);        while (num[i]--) printf("%d ",f[j++]);        putchar('\n');    }}