pat甲1021.Deepest Root（简单dfs）

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 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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

tips:

题意是找出一个树中使得树的深度最大的根节点，如果有多个，则升序输出。

一道入门的搜索题了。注意到题目中给出点的数目多达10000个，使用邻接表是必要的

我是先深搜找出联通分量的个数，如果是一个森林，则直接输出连通分量的个数结束。

否则，分别对每一个点再次搜索，统计每一个点作为根的时候树的深度，用一个数组存储。

注意在每次搜索的时候将标记数组清空。

以上！

#include<iostream>#include<vector>#include<cstring>using namespace std;vector<int>graph[10002];//邻接表 int n,ans,_max;int flag[10002];//作为根时树的深度 int book[10002];//标记数组 void dfs1(int step){book[step]=1;for(int i=0;i<graph[step].size();i++){if(!book[graph[step][i]])dfs1(graph[step][i]);} }void dfs2(int step,int cnt){_max=max(_max,cnt);book[step]=1;for(int i=0;i<graph[step].size();i++){if(!book[graph[step][i]])dfs2(graph[step][i],cnt+1);} }int main(){//邻接表建图 cin>>n;for(int i=0;i<n-1;i++){int u,v;cin>>u>>v;graph[u].push_back(v);graph[v].push_back(u);}for(int i=1;i<=n;i++){if(!book[i]){dfs1(i);ans++;}} //如果联通分量的数目不唯一  if(ans>1){ cout<<"Error: "<<ans<<" components"<<endl; goto tag; } else { for(int i=1;i<=n;i++){ memset(book,0,sizeof(book)); _max=0; dfs2(i,1); flag[i]=_max;} for(int i=1;i<=n;i++){_max=max(_max,flag[i]);}for(int i=1;i<=n;i++){if(flag[i]==_max)cout<<i<<endl;} } tag:return 0;}