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 root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes' numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print "Error: K components" where K is the number of connected components in the graph.

Sample Input 1:

5
1 2
1 3
1 4
2 5

Sample Output 1:

3
4
5

Sample Input 2:

5
1 3
1 4
2 5
3 4

Sample Output 2:

Error: 2 components

题目大意:对于图如果其是联通无环的,那么就是树,找出最深的树的所有起点。

//不知道怎么找到所有的,找到两个可还行。

//使用并查集判断是否是树,然后呢?

代码来自:https://www.liuchuo.net/archives/2348

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int n, maxheight = ;
vector<vector<int>> v;
bool visit[];
set<int> s;
vector<int> temp;
void dfs(int node, int height) {
if(height > maxheight) {
temp.clear();//只存放最深的。
temp.push_back(node);
maxheight = height;
} else if(height == maxheight){
temp.push_back(node);
}
visit[node] = true;//一般图访问点完了之后,都要标记的,防止其下一个点再返回去访问它。
for(int i = ; i < v[node].size(); i++) {
if(visit[v[node][i]] == false)
dfs(v[node][i], height + );
}
}
int main() {
scanf("%d", &n);
v.resize(n + );
int a, b, cnt = , s1 = ;
for(int i = ; i < n - ; i++) {
scanf("%d%d", &a, &b);
v[a].push_back(b);//使用二维向量来存储。
v[b].push_back(a);
}
for(int i = ; i <= n; i++) {
if(visit[i] == false) {
dfs(i, );
if(i == ) {//其实这里随便一个点均可。
if (temp.size() != ) s1 = temp[];
for(int j = ; j < temp.size(); j++)
s.insert(temp[j]);//再使用集合存储,防止重复。
}
cnt++;
}
}
if(cnt >= ) {
printf("Error: %d components", cnt);
} else {
temp.clear();
maxheight = ;
fill(visit, visit + , false);
dfs(s1, );
for(int i = ; i < temp.size(); i++)
s.insert(temp[i]);
for(auto it = s.begin(); it != s.end(); it++)
printf("%d\n", *it);
}
return ;
}

//厉害,使用dfs,传递层数参数,并且有一个maxHeight来保存最大的,厉害。

PAT 1021 Deepest Root[并查集、dfs][难]的更多相关文章

  1. 1021.Deepest Root (并查集+DFS树的深度)

    A graph which is connected and acyclic can be considered a tree. The height of the tree depends on t ...

  2. PAT 甲级 1021 Deepest Root (并查集,树的遍历)

    1021. Deepest Root (25) 时间限制 1500 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A graph ...

  3. [PAT] 1021 Deepest Root (25)(25 分)

    1021 Deepest Root (25)(25 分)A graph which is connected and acyclic can be considered a tree. The hei ...

  4. PAT 1021 Deepest Root

    #include <cstdio> #include <cstdlib> #include <vector> using namespace std; class ...

  5. PAT甲级1021. Deepest Root

    PAT甲级1021. Deepest Root 题意: 连接和非循环的图可以被认为是一棵树.树的高度取决于所选的根.现在你应该找到导致最高树的根.这样的根称为最深根. 输入规格: 每个输入文件包含一个 ...

  6. PAT 甲级 1021 Deepest Root (25 分)(bfs求树高,又可能存在part数part>2的情况)

    1021 Deepest Root (25 分)   A graph which is connected and acyclic can be considered a tree. The heig ...

  7. PAT甲题题解-1021. Deepest Root (25)-dfs+并查集

    dfs求最大层数并查集求连通个数 #include <iostream> #include <cstdio> #include <algorithm> #inclu ...

  8. 1021. Deepest Root (25)——DFS+并查集

    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 deep ...

  9. 1021. Deepest Root (25) -并查集判树 -BFS求深度

    题目如下: A graph which is connected and acyclic can be considered a tree. The height of the tree depend ...

随机推荐

  1. 【大数据系列】hadoop核心组件-MapReduce

    一.引入 hadoop的分布式计算框架(MapReduce是离线计算框架) 二.MapReduce设计理念 移动计算,而不是移动数据. Input HDFS先进行处理切成数据块(split)   ma ...

  2. 【分享】IT产业中的三大定理(三) —— 反摩尔定理 (Reverse Moore's Law)

    Google(谷歌)的 CEO 埃里克·施密特在一次采访中指出,如果你反过来看摩尔定理,一个 IT 公司如果今天和十八个月前卖掉同样多的.同样的产品,它的营业额就要降一半.IT 界把它称为反摩尔定理. ...

  3. Cross-compilation using Clang

    Introduction This document will guide you in choosing the right Clang options for cross-compiling yo ...

  4. rman 中遇到 ORA-01861

    RMAN> run{ 2> sql 'alter session set nls_date_format="yyyy-mm-dd hh24:mi:ss"'; 3> ...

  5. Word 2013无法启用Restrict Editing解决方法

    当前文档可能是Mail Merge Letter type document,MAILINGS -> Start Mail Merge -> Normal Word Document保存即 ...

  6. SIM900A基站定位调试笔记 -转

    第1步:ATE1 握手并设置回显 第2步:AT+CGMR 查看SIM900的版本信号 第3步:AT+CSQ 查看信号质量 第4步:AT+CREG? 查看GSM是否注册成功 第5步:AT+CGREG?  ...

  7. Linux渗透之反弹Shell

    前言 当我们在渗透Linux主机时,反弹一个交互的shell是非常有必要的.在搜索引擎上搜索关键字“Linux 反弹shell”,会出现一大堆相关文章,但是其内容不但雷同,而且都仅仅是告诉我们执行这个 ...

  8. 【LOJ6077】「2017 山东一轮集训 Day7」逆序对 生成函数+组合数+DP

    [LOJ6077]「2017 山东一轮集训 Day7」逆序对 题目描述 给定 n,k ,请求出长度为 n的逆序对数恰好为 k 的排列的个数.答案对 109+7 取模. 对于一个长度为 n 的排列 p ...

  9. [工具] multidesk

    MultiDesk 是一个选项卡(TAB标签)方式的远程桌面连接 (Terminal Services Client). http://www.hoowi.com/multidesk/index_ch ...

  10. Java虚拟机九 java.lang.String在虚拟机中的实现

    在Java中,Java的设计者对String对象进行了大量的优化,主要有三个特点: 1.不变性: 不变性是指String对象一旦生成,则不能再对它进行改变.String的这个特点可以泛化成不变(imm ...