1021 Deepest Root (25)(25 分)A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest…

1021 Deepest Root (25 分)   A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest r…

http://pat.zju.edu.cn/contests/pat-a-practise/1021 无环连通图也可以视为一棵树,选定图中任意一点作为根,如果这时候整个树的深度最大,则称其为 deepest root. 给定一个图,按升序输出所有 deepest root.如果给定的图有多个连通分量,则输出连通分量的数量. 1.使用并查集判断图是否为连通的. 2.任意选取一点,做 dfs 搜索,选取其中一个最远距离的点 A,再做一次 dfs,找到的所有距离最远的点以及点 A 都是 deepest…

题目如下: A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root. Input Specificat…

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root. Input Specification: E…

problem A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root. Input Specific…

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root. Input Specification: E…

先并查集判断连通性,然后暴力每个点作为根节点判即可. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<queue> #include<vector> using namespace std; struct Edge { int a,b; }e[]; ],dep,U; ]; v…

dfs求最大层数并查集求连通个数 #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <string> #include <vector> using namespace std; /* dfs求最大层数 并查集求连通个数 */ +; int n; ; //最大层数 vector<int>deepro…

题意: 输入一个正整数N(N<=10000),然后输入N-1条边,求使得这棵树深度最大的根节点,递增序输出.如果不是一棵树,输出这张图有几个部分. trick: 时间比较充裕数据可能也不是很极限,所以用了一个暴力的手段.直接枚举度数为1的点进行深度优先搜索,如果递归了一万次还没有return就当它有环.(这道题的数据不存在只有一个带有自环的部分,即如有自环图必定不止一个)正解应当是dfs一次以后找到一些可能是最深根的叶子结点们,然后对这些叶子节点们依次dfs,再找到对应的一些叶子节点们.成对的叶…