A graph which is connected and acyclic can be considered a tree. The hight 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 (≤10​^4​​) 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

题目大意

一个连通的无环图可以看做一个树，而树的高度取决于根节点的选择。题目要求找到使树最高的根，如符合条件的根不唯一则找到所有这样的根并按增序输出。如所给图并不连通，则输出其连通子图数量。

解题思路

1. 用查并集检查所有结点组成多少联通子图，同时用邻接表保存图，如子图数目大于1则输出子图数量并返回0值；
2. 如为连通图，随便从一个节点出发进行DFS 找到最深的若干个节点；
3. 从某个最深节点出发进行DFS找到最深节点，结合第二步所得即为结果；
4. 返回所求结果并返回0值。

代码

#include<stdio.h>
#include<vector>
using namespace std;
#define maxn 10010

int Deepest[maxn];
int numDeep,maxDeep=-1;

vector<int> Adj[maxn];
int father[maxn];
int isroot[maxn];
int isdeep[maxn];

int n;
int findFather(int x){
    int z,a;
    a=x;
    while(father[x]!=x){
        x=father[x];
    }
    while(a!=x){
        z=a;
        a=father[a];
        father[z]=x;
    }
    return x;
}

void Union(int a,int b){
    int fa,fb;
    fa=findFather(a);
    fb=findFather(b);
    if(fa!=fb){
        father[fa]=fb;
    }
}

void Init(){
    int i,a,b;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        father[i]=i;
        isroot[i]=0;
        isdeep[i]=0;
    }
    for(i=1;i<n;i++){
        scanf("%d%d",&a,&b);
        Adj[a].push_back(b);
        Adj[b].push_back(a);
        Union(a,b);
    }
    return;
}

void DFS(int u,int depth,int pre){
    int i,v;
    if(depth>maxDeep){
        maxDeep=depth;
        numDeep=0;
        Deepest[numDeep++]=u;
    }else if(depth==maxDeep){
        Deepest[numDeep++]=u;
    }
    for(i=0;i<Adj[u].size();i++){
        v=Adj[u][i];
        if(v!=pre){
            DFS(v,depth+1,u);
        }
    }
    return;
}


int main(){
    int nnum=0,i; 
    Init();
    for(i=1;i<=n;i++){
        isroot[findFather(i)]=1;
    }
    for(i=1;i<=n;i++){
        nnum+=isroot[i];
    }
    if(nnum>1){
        printf("Error: %d components\n",nnum);
    }else{
        DFS(1,0,0);
        for(i=0;i<numDeep;i++){
            isdeep[Deepest[i]]=1;
        }
        DFS(Deepest[0],0,0);
        for(i=0;i<numDeep;i++){
            isdeep[Deepest[i]]=1;
        }
        for(i=1;i<=n;i++){
            if(isdeep[i]){
                printf("%d\n",i);
            }
        }
    }
    return 0;
}

运行结果