PAT 1021. Deepest Root (25)

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

DFS一次，选最远点再次DFS,并集就是所求。

#include <cstdio>#include<string.h>#include<string>#include<map>#include<vector>#include<stdio.h>#include<set>#include<iostream>#include <algorithm>using namespace std;vector<int> edge[10005];int mark[10005]={0};int deepest=-1;int path[10005];int p=0;int tmp;void dfs(int x,int deep){    mark[x]=1;    if(deep>deepest)    {        deepest=deep;        path[0]=x;        p=1;    }    else if(deep==deepest)    {        path[p++]=x;    }    for(int i=0;i<edge[x].size();i++)    {        if(mark[edge[x][i]]==0) dfs(edge[x][i],deep+1);    }}int main(){    int n;    cin>>n;   for(int i=1;i<n;i++)   {       int x,y;       scanf("%d %d",&x,&y);       edge[x].push_back(y);       edge[y].push_back(x);   }     dfs(1,0);     int flag=0;     for(int i=1;i<=n;i++)       if(mark[i]==0) {flag=1;break;}     if(flag==1)     {         int cnt=1;         for(int i=1;i<=n;i++)          if(mark[i]==0)         {             cnt++;dfs(i,0);         }         printf("Error: %d components",cnt);     }     else     {         tmp=path[p-1];         set<int> S;         for(int i=0;i<p;i++)         S.insert(path[i]);         p=0;         memset(mark,0,sizeof(mark));         dfs(tmp,0);         for(int i=0;i<p;i++)         S.insert(path[i]);         for(set<int>::iterator it=S.begin();it!=S.end();it++)         {             cout<<*it<<endl;         }     }    return 0;}