Codeforces Round #403 (Div. 2) C. Andryusha and Colored Balloons DFS

C. Andryusha and Colored Balloons

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

32 31 3

output

31 3 2 

input

52 35 34 31 3

output

51 3 2 5 4 

input

52 13 24 35 4

output

31 2 3 1 2 

Note

In the first sample the park consists of three squares: 1 → 3 → 2. Thus, the balloon colors have to be distinct.

Illustration for the first sample.

In the second example there are following triples of consequently connected squares:

• 1 → 3 → 2
• 1 → 3 → 4
• 1 → 3 → 5
• 2 → 3 → 4
• 2 → 3 → 5
• 4 → 3 → 5

We can see that each pair of squares is encountered in some triple, so all colors have to be distinct.Illustration for the second sample.

In the third example there are following triples:

• 1 → 2 → 3
• 2 → 3 → 4
• 3 → 4 → 5

We can see that one or two colors is not enough, but there is an answer that uses three colors only.Illustration for the third sample.

Source

Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals)

My Solution

题意：给出一颗无根树，要求如果a-b相连 b-c相连，则要求abc涂上不同的颜色，要求用最少的颜色给这颗树上色且求具体涂色。

DFS

首先这个最少的颜色数必定是 最大度+1，

然后具体涂色的时候可以dfs来涂色，因为dfs的时候孙子节点的颜色只跟相应子节点和父节点有关，跟该父节点的其它子节点没有关系，

所以用dfs来涂色。

复杂度 O(n)

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;typedef long long LL;const int maxn = 2e5 + 8;vector<int> sons[maxn];int p[maxn];void dfs(const int &g, const int &u){        int j, sz, ptr = 1;        sz = sons[u].size();        if(sz == 1) return;        for(j = 0; j < sz; j++){            while(ptr == p[g] || ptr == p[u]) ptr++;            if(sons[u][j] == g) continue;            p[sons[u][j]] = ptr; ptr++;            dfs(u, sons[u][j]);        }}int main(){    #ifdef LOCAL    freopen("c.txt", "r", stdin);    //freopen("c.out", "w", stdout);    int T = 4;    while(T--){    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int n, v, u, t, ans = 0, i, j, sz, maxi;    cin >> n; t = n - 1;    while(t--){        cin >> u >> v;        sons[u].push_back(v);        sons[v].push_back(u);    }    for(i = 1; i <= n; i++){        sz = sons[i].size();        if(ans < sz + 1){            ans = sz + 1;            maxi = i;        }    }    p[maxi] = ans; p[0] = 0;    dfs(0, maxi);    cout << ans << "\n" << p[1];    for(i = 2; i <= n; i++) cout << " " << p[i];    #ifdef LOCAL    for(int i = 1; i <= n; i++){        sons[i].clear();    }    memset(p, 0, sizeof p);    cout << endl;    }    #endif // LOCAL    return 0;}

                                                                                                                                             ------from ProLights