1021. Deepest Root (25)(图的直径）

1. 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

#include<iostream>#include<vector>#include<set>#include<cstdio>#include<cstring>using namespace std;int father[10005];int vis[10005];vector <int > v[10005];set<int> se;int maxlev=-1;int findfa(int x){    if (x==father[x]) return x;    else return father[x]=findfa(father[x]);}void dfs(int s,int leval){    if (leval>maxlev)    {        se.clear();        maxlev=leval;        se.insert(s);    }    else if (leval==maxlev)    {        se.insert(s);    }    for (int i=0;i<v[s].size();i++)    {        if (vis[v[s][i]]==0)        {            vis[v[s][i]]=1;            dfs(v[s][i],leval+1);             vis[v[s][i]]=0;        }    }}int main(){    int n,a,b,sum=0;    cin>>n;    for (int i=0;i<=n;i++)        father[i]=i;    for (int i=1;i<n;i++)    {        cin>>a>>b;        v[a].push_back(b);        v[b].push_back(a);        int fa=findfa(a);        int fb=findfa(b);        if(fa!=fb)        father[fa]=fb;    }    for (int i=1;i<=n;i++)        if (father[i]==i)            sum++;    if (sum>1)    {        printf("Error: %d components",sum);         return 0;    }    vis[1]=1;    dfs(1,1);    vis[1]=0;    set<int>ans;    for (set<int>::iterator it=se.begin();it!=se.end();it++)        ans.insert(*it);    int num=*se.begin();    maxlev=-1;    memset(vis,0,sizeof(vis));    se.clear();    dfs(num,1);    for (set<int>::iterator it=se.begin();it!=se.end();it++)        ans.insert(*it);    for (set<int>::iterator it=ans.begin();it!=ans.end();it++)        cout<<*it<<endl;    return 0;}