PAT1021:Deepest Root
1021. Deepest Root (25)
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 思路 二次dfs
1.第一次dfs确定图是否连通,如果连通找到最深的那个点first(多个就取最先被找到的),没有输出题目要求的错误信息。
2.第二次dfs重置所有状态,然后从first开始dfs,找到的所有的最深的点即是题目要求的节点,依次插入一个set容器中(每次插入会自动排序)。 3.输出set中的所有元素就行。 代码
#include<iostream>
#include<vector>
#include<set>
using namespace std;
vector<vector<int>> graph;
vector<int> highestNodes;
vector<bool> visits(10005,false);
int maxheight = 1;
set<int> results; void dfs(int root,int height)
{
visits[root] = true;
if(height >= maxheight)
{
if(height > maxheight)
{
highestNodes.clear();
maxheight = height;
}
highestNodes.push_back(root);
}
for(int i = 0;i < graph[root].size();i++)
{
if(!visits[graph[root][i]])
dfs(graph[root][i],height + 1);
}
} inline void resetVisits(const int n)
{
for(int i = 0;i <= n;i++)
visits[i] = false;
} int main()
{
int N;
while(cin >> N)
{
//input
graph.resize(N + 1);
for(int i = 1;i < N;i++)
{
int a,b;
cin >> a >> b;
graph[b].push_back(a);
graph[a].push_back(b);
} int cnt = 0;
//handle
int first = -1; //最高的节点之一
for(int i = 1;i <= N;i++)
{
if(!visits[i])
{
cnt++;
dfs(i,1);
if( i == 1)
{
for(int j = 0;j < highestNodes.size();j++)
{
results.insert(highestNodes[j]);
if(j == 0)
first = highestNodes[j];
}
}
}
} if(cnt > 1)
cout << "Error: "<< cnt <<" components" << endl;
else
{
highestNodes.clear();
maxheight = 1;
resetVisits(N);
dfs(first,1);
for(int i = 0;i < highestNodes.size();i++ )
results.insert(highestNodes[i]);
for(auto it = results.begin();it != results.end();it++)
cout << *it << endl;
}
}
}
PAT1021:Deepest Root的更多相关文章
- PAT-1021 Deepest Root (25 分) 并查集判断成环和联通+求树的深度
A graph which is connected and acyclic can be considered a tree. The height of the tree depends on t ...
- PAT1021. Deepest Root (25)
之前不知道怎么判断是不是树,参考了 http://blog.csdn.net/eli850934234/article/details/8926263 但是最后有一个测试点有超时,在bfs里我用了数组 ...
- 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 ...
- 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 ...
- 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 ...
- A1021. Deepest Root
A graph which is connected and acyclic can be considered a tree. The height of the tree depends on t ...
- PAT A1021 Deepest Root (25 分)——图的BFS,DFS
A graph which is connected and acyclic can be considered a tree. The hight of the tree depends on th ...
- 1021. Deepest Root (25)
A graph which is connected and acyclic can be considered a tree. The height of the tree depends on t ...
- PAT 甲级 1021 Deepest Root
https://pintia.cn/problem-sets/994805342720868352/problems/994805482919673856 A graph which is conne ...
随机推荐
- 关于学习MMU的一点感想
MMU的一个主要服务是能把各个人物作为各自独立的程序在其自己的虚拟存储空间中运行. 虚拟存储器系统的一个重要特征是地址重定位.地址重定位是将处理器核产生的地址转换到主存的不同地址,转换由MMU硬件完成 ...
- cocos2d-x 游戏开发之有限状态机(FSM) (二)
cocos2d-x 游戏开发之有限状态机(FSM) (二) 1 状态模式
- 如何搭建modem编译环境
[DESCRIPTION] (1)MT6577以及之前的chip平台(如MT6575,73等) 的modem编译环境和MTK的Feature Phone的编译环境一样,即Windows+RVCT (2 ...
- PhotoShop 图像处理 算法 汇总
不定期更新 ...... 直接点标题即可链接到原文. OpenCV 版:OpenCV 图像处理 图层混合算法: PS图层混合算法之一(不透明度,正片叠底,颜色加深,颜色减淡)PS图层混合算法之二(线性 ...
- 色彩转换——RGB & HSI
RGB to HSI I=(R+G+B)/3; S=1-3*min(R,G,B)/(R+G+B); H = cos^(-1)((0.5*((R-G)+(R-B))) / ((R-G)^2 + (R-B ...
- LeetCode(64)- Min Stack
题目: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time ...
- ruby中__FILE__,$FILENAME,$PROGRAM_NAME,$0等类似变量的含义
ruby中有4个类似的变量(常量),他们分别是: __FILE__,$FILENAME,$PROGRAM_NAME,$0 他们分别在代码中表示神马呢?我们用实际的例子说明一下: x.rb #!/usr ...
- 图片验证码demo示例
1.首先我们需要一个生成图片验证码图片的一个工具类(下方会有代码示例) 代码如下: package com.util; import java.awt.BasicStroke; import java ...
- ASP.NET MVC不可或缺的部分——DI(IOC)容器及控制器重构的剖析(DI的实现原理)
IoC框架最本质的东西:反射或者EMIT来实例化对象.然后我们可以加上缓存,或者一些策略来控制对象的生命周期,比如是否是单例对象还是每次都生成一个新的对象. DI实现其实很简单,首先设计类来实现接口, ...
- 二叉树的建立以及遍历的多种实现(python版)
二叉树是很重要的数据结构,在面试还是日常开发中都是很重要的角色. 首先是建立树的过程,对比C或是C++的实现来讲,其涉及到了较为复杂的指针操作,但是在面向对象的语言中,就不需要考虑指针, 内存等.首先 ...