【Codeforces Round #403】Codeforces 781A Andryusha and Colored Balloons

Andryusha goes through a park each day. The squares and paths between
them look boring to Andryusha, so he decided to decorate them.

The park consists of n squares connected with (n - 1) bidirectional
paths in such a way that any square is reachable from any other using
these paths. Andryusha decided to hang a colored balloon at each of
the squares. The baloons’ colors are described by positive integers,
starting from 1. In order to make the park varicolored, Andryusha
wants to choose the colors in a special way. More precisely, he wants
to use such colors that if a, b and c are distinct squares that a and
b have a direct path between them, and b and c have a direct path
between them, then balloon colors on these three squares are distinct.

Andryusha wants to use as little different colors as possible. Help
him to choose the colors! Input

The first line contains single integer n (3 ≤ n ≤ 2·105) — the number
of squares in the park.

Each of the next (n - 1) lines contains two integers x and y
(1 ≤ x, y ≤ n) — the indices of two squares directly connected by a
path.

It is guaranteed that any square is reachable from any other using the
paths. Output

In the first line print single integer k — the minimum number of
colors Andryusha has to use.

In the second line print n integers, the i-th of them should be equal
to the balloon color on the i-th square. Each of these numbers should
be within range from 1 to k.

dfs一遍每次贪心选最小的。因为各个子树再往下就没有关系了，选别的也没有意义。

#include<cstdio>#include<algorithm>using namespace std;int fir[200010],ne[400010],to[400010],c[400010],n,m;void add(int num,int u,int v){    ne[num]=fir[u];    fir[u]=num;    to[num]=v;}void dfs(int u,int fa){    int v;    for (int i=fir[u],j=1;i;i=ne[i])        if ((v=to[i])!=fa)        {            while (j==c[u]||(fa!=-1&&j==c[fa])) j++;            m=max(m,j);            c[v]=j;            dfs(v,u);            j++;        }}int main(){    int u,v;    scanf("%d",&n);    for (int i=1;i<n;i++)    {        scanf("%d%d",&u,&v);        add(i*2,u,v);        add(i*2+1,v,u);    }    m=c[1]=1;    dfs(1,-1);    printf("%d\n",m);    for (int i=1;i<=n;i++) printf("%d ",c[i]);}