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…

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…

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点的颜色不同,只需考虑当前节点是否与父亲结点和父亲的父亲结点颜色是否相同. /** @Date : 2017-05-09 22:37:25 * @FileName: 782C DFS.cpp * @Platform: Windows * @Author : Lweleth (SoundEarlf@g…

地址:http://codeforces.com/contest/782/problem/C 题目: 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…

C - Andryusha and Colored Balloons 思路: 水题: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 200005 ],V[maxn<<],dis[maxn],ans; void dfs(int now,int fa) { ; for…

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…

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/转载请注明出处,侵权必究,保留最终解释权! 题目链接:codeforces781A Andryusha and Colored Balloons 正解:构造+结论 解题报告: 考虑答案显然是$max(度数)+1$,这个似乎很好想,对于每个点我需要保证与之相邻的所有点颜色互不相同且与自己不同,那么需要…

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…

http://codeforces.com/contest/782/problem/C 题意:给一棵树染最少的颜色,使得相邻距离为2的点都是不同的颜色,问最少是多少种颜色并输出每个点的颜色. 思路:比赛的时候没想到是找度最大的一个点并+1就是总颜色数,一直想怎么构造. 最后的总颜色数是度最大的一个点的度数+1.因为我们选的这个点到儿子的距离为1,因此其某一个儿子到另一个儿子的距离为2,就是这些儿子都要染成不同的颜色,+1是因为自己本身也要是不同的颜色. 然后确定了总颜色数,就可以DFS染色了.…

题目链接:http://codeforces.com/contest/742/problem/E 题意: 有一个环形的桌子,一共有n对情侣,2n个人,一共有两种菜. 现在让你输出一种方案,满足以下要求: 情侣间吃不同的菜 相邻的三个人不能都吃同一种菜 输出任意一个解: 先将相邻的两个人连边,这样就满足了3个人不吃同样一种菜. 情侣间连边. 图中就不存在奇数环. 那么就一定存在解.然后DFS染色就OK 了. #include <bits/stdc++.h> using namespace std…