C. Andryusha and Colored Balloons
2 seconds
256 megabytes
standard input
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!
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.
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.
3
2 3
1 3
3
1 3 2
5
2 3
5 3
4 3
1 3
5
1 3 2 5 4
5
2 1
3 2
4 3
5 4
3
1 2 3 1 2
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
- 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 ...
- 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 ...
- 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 ...
- 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/ ...
- 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 ...
- 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, ...
- CF781A Andryusha and Colored Balloons
题意: Andryusha goes through a park each day. The squares and paths between them look boring to Andryu ...
- 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 ...
- 【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
- css系列,选择器权重计算方式
CSS选择器分基本选择器(元素选择器,类选择器,通配符选择器,ID选择器,关系选择器), 属性选择器,伪类选择器,伪元素选择器,以及一些特殊选择器,如has,not等. 在CSS中,权重决定了哪些CS ...
- Java事务与JTA
一.什么是JAVA事务 通俗的理解,事务是一组原子操作单元,从数据库角度说,就是一组SQL指令,要么全部执行成功,若因为某个原因其中一条指令执行有错误,则撤销先前执行过的所有指令.更简答的说就是:要么 ...
- [学习总结]7、Android AsyncTask完全解析,带你从源码的角度彻底理解
我们都知道,Android UI是线程不安全的,如果想要在子线程里进行UI操作,就需要借助Android的异步消息处理机制.之前我也写过了一篇文章从源码层面分析了Android的异步消息处理机制,感兴 ...
- Hibernate 总结(转)
JMX:Java Management Extensions.JCA: J2EE Contector ArchitectureJNDI: Java Namind and Directory Inter ...
- 【编程思想】【设计模式】【创建模式creational】原形模式Prototype
Python版 https://github.com/faif/python-patterns/blob/master/creational/prototype.py #!/usr/bin/env p ...
- 用运oracel中的伪列rownum分页
在实际应用中我们经常碰到这样的问题,比如一张表比较大,我们只要其中的查看其中的前几条数据,或者对分页处理数据.在这些情况下我们都需要用到rownum.因此我们要理解rownum的原理和使用方法. Or ...
- Dubbo提供者的异步执行
从前面"对提供者的异步调用"例子可以看出,消费者对提供者实现了异步调用,消费者线程的执行过程不再发生阻塞,但提供者对IO耗时操作仍采用的是同步调用,即IO操作仍会阻塞Dubbo的提 ...
- get请求url参数中有+、空格、=、%、&、#等特殊符号的问题解决
url出现了有+,空格,/,?,%,#,&,=等特殊符号的时候,可能在服务器端无法获得正确的参数值,如何是好?解决办法将这些字符转化成服务器可以识别的字符,对应关系如下:URL字符转义 用其它 ...
- 【阿菜做实践】利用ganache-cli本地fork以太坊主链分叉
前言 Fork主网意思是模拟具有与主网相同的状态的网络,但它将作为本地开发网络工作. 这样你就可以与部署的协议进行交互,并在本地测试复杂的交互.不用担心分叉主网作为测试链会占很多内存.这些方法都不会将 ...
- Linux下安装Calibre
目录 一.介绍 二.安装 三.测试 四.报错处理 一.介绍 Calibre是基于python的电子书制作软件,可导出PDF.EPUB.MOBI.Word格式电子书. 二.安装 yum -y insta ...