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 root。

考虑到为稀疏图,则使用动态链表

#include<stdio.h>
#include<stack>
#include<iostream>
#include<vector>
#include<set>
#include<queue>
using namespace std; vector<int> map[];
int f[];
int find(int k){
if(f[k]==-)return k;
else
return f[k]=find(f[k]);
} int um(int a,int b){
int fa,fb;
fa=find(a);
fb=find(b); if(fa==fb)return ;//表示有环 if(fa!=fb)
f[fa]=fb;
return ;
} int step[]; void dfs(int x,int STEP){ int i;
for(i=;i<map[x].size();i++){
if(step[map[x][i]]!=)continue;
step[map[x][i]]=STEP;
dfs(map[x][i],STEP+);
}
} int main()
{
int n;
while(scanf("%d",&n)!=EOF){
int i,a,b; for(i=;i<=n;i++){
f[i]=-;
step[i]=;
} int huan=;
for(i=;i<n;i++){
scanf("%d%d",&a,&b);
if(um(a,b)==)huan=; map[a].push_back(b);
map[b].push_back(a);
}
int connectAdd=;
set<int>set1;
int ttemp;
for(i=;i<=n;i++){
ttemp=find(i);
set1.insert(ttemp);
} if(set1.size()>=||huan==){
printf("Error: %d components\n",set1.size());continue;
}
step[]=;
dfs(,); int max=,ri;
for(i=;i<=n;i++){
if(max<step[i]){
max=step[i];
ri=i;
}
} for(i=;i<=n;i++){
step[i]=;
}
step[ri]=;
dfs(ri,); max=;
for(i=;i<=n;i++){
if(max<step[i]){
max=step[i];
}
} for(i=;i<=n;i++){
if(step[i]==max||step[i]==){
printf("%d\n",i);
}
}
} return ;
}

其实上面的算法还有点问题,虽然AC了

考虑

5
1 2
1 3
2 4
2 5

应该是输出

3

4

5

算法改进 以(第2次dfs最深的点)为起点,再做DFS得到的最深的点,这些点才是所有最深的点

#include<stdio.h>
#include<stack>
#include<iostream>
#include<vector>
#include<set>
#include<queue>
using namespace std; vector<int> map[];
int f[];
int find(int k){
if(f[k]==-)return k;
else
return f[k]=find(f[k]);
} int um(int a,int b){
int fa,fb;
fa=find(a);
fb=find(b); if(fa==fb)return ;//表示有环 if(fa!=fb)
f[fa]=fb;
return ;
} int step[];
int deepF[];//第2次dfs最深的点
int deepR[];//以(第2次dfs最深的点)为起点,做DFS得到的最深的点 void dfs(int x,int STEP){ int i;
for(i=;i<map[x].size();i++){
if(step[map[x][i]]!=)continue;
step[map[x][i]]=STEP;
dfs(map[x][i],STEP+);
}
} int main()
{
int n;
while(scanf("%d",&n)!=EOF){
int i,a,b,j; for(i=;i<=n;i++){
f[i]=-;
step[i]=;
deepF[i]=;
deepR[i]=;
} int huan=;
for(i=;i<n;i++){
scanf("%d%d",&a,&b);
if(um(a,b)==)huan=; map[a].push_back(b);
map[b].push_back(a);
}
int connectAdd=;
set<int>set1;
int ttemp;
for(i=;i<=n;i++){
ttemp=find(i);
set1.insert(ttemp);
} if(set1.size()>=||huan==){
printf("Error: %d components\n",set1.size());continue;
}
step[]=;
dfs(,); int max=,ri;
for(i=;i<=n;i++){
if(max<step[i]){
max=step[i];
ri=i;
}
} for(i=;i<=n;i++){
step[i]=;
}
step[ri]=;
dfs(ri,); max=;
for(i=;i<=n;i++){
if(max<step[i]){
max=step[i];
}
} for(i=;i<=n;i++){
if(step[i]==max||step[i]==){
//printf("%d\n",i);
deepF[i]=;
deepR[i]=;
}
} for(i=;i<=n;i++){
if(deepF[i]==)continue;
for(j=;j<=n;j++)step[j]=; dfs(i,);
for(j=;j<=n;j++){
if(step[j]==max)deepR[j]=;
}
} for(i=;i<=n;i++){
if(deepR[i]==)printf("%d\n",i);
}
} return ;
}

1021. Deepest Root (25)——DFS+并查集的更多相关文章

  1. 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 ...

  2. 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 ...

  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 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 ...

  5. 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 ...

  6. 1021. Deepest Root (25)

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

  7. PAT (Advanced Level) 1021. Deepest Root (25)

    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> #i ...

  8. 1021 Deepest Root (25)(25 point(s))

    problem A graph which is connected and acyclic can be considered a tree. The height of the tree depe ...

  9. 1021 Deepest Root (25 分)

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

随机推荐

  1. eclipse——添加Tomcat7.0服务器

    首先要安装好Tomcat 然后在eclipse中添加Tomcat 步骤如下 详细可参考这篇博客https://blog.csdn.net/u014079773/article/details/5139 ...

  2. PHP 开发环境搭建

    1. PHP (1) download PHP and extra the zip file to the folder “C:\tools\php” (2) add the path “;C:\to ...

  3. Count and Say,统计并输出,利用递归,和斐波那契数列原理一样。

    问题描述:n=1,返回“1”:n=2,返回“11”:n=3,返回“21”:n=4,返回1211,.... 算法分析:和斐波那契数列道理差不多,都是后一个要依赖前一个元素.因此可以使用递归,也可以使用迭 ...

  4. Treflection02_getMethods()_getMethod()

    1. package reflectionZ; import java.lang.reflect.Constructor; import java.lang.reflect.Method; impor ...

  5. 安装Charles报错

    去年用的是charles4.1.2版本,今年这个版本的安装包始终安装报错,不管公司电脑还是自己电脑........ 我的解决方案很Lower的.......... 登录Charles官网:https: ...

  6. request获取路径方式

    从request获取各种路径总结 request.getRealPath("url"); // 虚拟目录映射为实际目录 request.getRealPath("./&q ...

  7. 三十七 Python分布式爬虫打造搜索引擎Scrapy精讲—将bloomfilter(布隆过滤器)集成到scrapy-redis中

    Python分布式爬虫打造搜索引擎Scrapy精讲—将bloomfilter(布隆过滤器)集成到scrapy-redis中,判断URL是否重复 布隆过滤器(Bloom Filter)详解 基本概念 如 ...

  8. 远程登录MySQL

    mysql 远程连接数据库的二种方法   一.连接远程数据库: 1.显示密码 如:MySQL 连接远程数据库(192.168.5.116),端口“3306”,用户名为“root”,密码“123456” ...

  9. JSON和list之间的转换

    谷歌的Gson.jar: //list转换为json Gson gson = new Gson(); List<Person> persons = new ArrayList<Per ...

  10. ios上传图片遇见了一个TimeoutError(DOM Exception 23)异常

    TimeoutError(DOM Exception 23):The operation timed out 百度了下,没发现解决办法