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 ab 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
3
2 3
1 3
output
3
1 3 2
input
5
2 3
5 3
4 3
1 3
output
5
1 3 2 5 4
input
5
2 1
3 2
4 3
5 4
output
3
1 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.

思路:dfs;

每个节点只会由他的父亲节点,祖父节点,还有兄弟节点来影响,那么dfs,然后每层从1开始找不是父亲节点,祖父节点,还有兄  弟节值的最小值,复杂度O(n);

 1 #include<iostream>
2 #include<stdlib.h>
3 #include<queue>
4 #include<string.h>
5 #include<stdio.h>
6 #include<stack>
7 #include<vector>
8 using namespace std;
9 vector<int>vec[1000000];
10 int minn = 0;
11 int ask[1000000];
12 void dfs(int n,int fa);
13 int main(void)
14 {
15 int n;
16 scanf("%d",&n);
17 for(int i = 0; i <n-1; i++)
18 {
19 int x,y;
20 scanf("%d %d",&x,&y);
21 vec[x].push_back(y);
22 vec[y].push_back(x);
23 }ask[1] = 1;
24 dfs(1,0);
25 printf("%d\n",minn);
26 printf("%d",ask[1]);
27 for(int i = 2; i <= n; i++)
28 printf(" %d",ask[i]);
29 printf("\n");
30 return 0;
31 }
32 void dfs(int n,int fa)
33 {
34 int cnt = 1;
35 for(int i = 0; i <vec[n].size(); i++)
36 {
37 int ic = vec[n][i];
38 if(ic != fa)
39 {
40 while(cnt == ask[fa]||cnt == ask[n])
41 {
42 cnt++;
43 }
44
45 ask[ic] = cnt++;
46 dfs(ic,n);
47 }
48 }
49 minn = max(minn,cnt-1);
50 }

C. Andryusha and Colored Balloons的更多相关文章

  1. code force 403C.C. Andryusha and Colored Balloons

    C. Andryusha and Colored Balloons time limit per test 2 seconds memory limit per test 256 megabytes ...

  2. Codeforces 782C. Andryusha and Colored Balloons 搜索

    C. Andryusha and Colored Balloons time limit per test:2 seconds memory limit per test:256 megabytes ...

  3. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) C Andryusha and Colored Balloons

    地址:http://codeforces.com/contest/782/problem/C 题目: C. Andryusha and Colored Balloons time limit per ...

  4. codeforces781A Andryusha and Colored Balloons

    本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/ ...

  5. AC日记——Andryusha and Colored Balloons codeforces 780c

    C - Andryusha and Colored Balloons 思路: 水题: 代码: #include <cstdio> #include <cstring> #inc ...

  6. CodeForces - 780C Andryusha and Colored Balloons(dfs染色)

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

  7. CF781A Andryusha and Colored Balloons

    题意: Andryusha goes through a park each day. The squares and paths between them look boring to Andryu ...

  8. 782C. Andryusha and Colored Balloons DFS

    Link 题意: 给出一棵树,要求为其染色,并且使任意节点都不与距离2以下的节点颜色相同 思路: 直接DFS.由某节点出发的DFS序列,对于其个儿子的cnt数+1,那么因为DFS遍历的性质可保证兄弟结 ...

  9. 【贪心】【DFS】Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) C. Andryusha and Colored Balloons

    从任意点出发,贪心染色即可. #include<cstdio> #include<algorithm> using namespace std; int v[200010< ...

随机推荐

  1. kubernetes部署 flannel网络组件

    创建 flannel 证书和私钥flannel 从 etcd 集群存取网段分配信息,而 etcd 集群启用了双向 x509 证书认证,所以需要为 flanneld 生成证书和私钥. cat > ...

  2. 【Redis集群原理专题】分析一下相关的Redis集群模式下的脑裂问题!

    技术格言 世界上并没有完美的程序,但是我们并不因此而沮丧,因为写程序就是一个不断追求完美的过程. 什么是脑裂 字面含义 首先,脑裂从字面上理解就是脑袋裂开了,就是思想分家了,就是有了两个山头,就是有了 ...

  3. Spring Security 基于URL的权限判断

    1.  FilterSecurityInterceptor 源码阅读 org.springframework.security.web.access.intercept.FilterSecurityI ...

  4. java四则运算规则

    java四则运算规则 1.基本规则 运算符:进行特定操作的符号.例如:+ 表达式:用运算符连起来的式子叫做表达式.例如:20 + 5.又例如:a + b 四则运算: 加:+ 减:- 乘:* 除:/ 取 ...

  5. 学习Java的第十八天

    一.今日收获 1.java完全学习手册第三章算法的3.1比较值 2.看哔哩哔哩上的教学视频 二.今日问题 1.在第一个最大值程序运行时经常报错. 2.哔哩哔哩教学视频的一些术语不太理解,还需要了解 三 ...

  6. Linux 参数代换 命令 xargs

    xargs 命令也是管道命令中的一员.xargs命令的功能简单来说就是参数代换.那么什么叫做参数代换,这里首先要了解管道的概念.在 linux管道 命令一节中我们详细介绍了管道命令的概念.这里我们只是 ...

  7. Shell 打印空行的行号

    目录 Shell 打印空行的行号 题解 Shell 打印空行的行号 写一个 bash脚本以输出一个文本文件 nowcoder.txt中空行的行号,可能连续,从1开始 示例: 假设 nowcoder.t ...

  8. Redis6 新特性

    Redis6新特性 ACL安全策略 ACL(access control list): 访问控制列表,可以设置多个用户,并且给每个用户单独设置命令权限和数据权限 default用户和使用require ...

  9. js 时间戳转换为年月日时分秒的格式

    <script type="text/javascript"> var strDate = ''; $(function(){ // 获取时间戳 var nowDate ...

  10. 【C/C++】旋转数组的最小数字/ 剑指offer

    #include <bits/stdc++.h> using namespace std; class Solution { public: int minNumberInRotateAr ...