1021 Deepest Root (25分)

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 (≤) 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

题意：
判断是否为一棵树，若不是，则输出连通分量个数；若是，则输出构成树最深高度时，树的根结点。
思路：
最深高度时树的根结点，其实就是找树中的最远距离路径上的起点和终点，因此需要先进行一遍dfs找到最远距离的n个起点，再从这些起点中任选一个dfs，找到最远距离的n个终点，起点和终点都是所求点。
那为什么第一次dfs找到的点就可以当作起点呢？可以用反证法，若选其他点做起点，它所在的深度不是最大，再利用例一的图进行dfs，发现不符合题意（可以多模拟几遍进行理解）。

AC代码（参考柳神的代码）

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<set>
#include<cmath>
#include<cstring>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn=10010;
vector<vector<int> > ve;//存储边
int vis[maxn];
vector<int> temp;//存储每一次遍历的根节点
set<int> s;//存储根节点
int n;
int maxdep=0;
void dfs(int node,int depth) {
    if(depth>maxdep) {
        maxdep=depth;
        temp.clear();//深度增加，则temp中存的已经不是最远的结点，需要清空
        temp.push_back(node);
    }
    else if(depth==maxdep) {
        temp.push_back(node);
    }
    vis[node]=1;
    for(int i=0; i<ve[node].size(); i++) {
        if(vis[ve[node][i]]==0) {//注意是ve[node][i]不是i
            dfs(ve[node][i],depth+1);
        }
    }
}

int main() {
    int a,b;
    scanf("%d",&n);
    ve.resize(n+1);
    fill(vis,vis+maxn,0);
    for(int i=1; i<n; i++) {
        scanf("%d %d",&a,&b);
        ve[a].push_back(b);
        ve[b].push_back(a);
    }
    int cnt=0;
    int s1;
    for(int i=1; i<=n; i++) {
        if(vis[i]==0) {
            dfs(i,1);
            if(i==1) {
                if(!temp.empty()) {
                    s1=temp[0];
                    for(int i=0; i<temp.size(); i++) {
                        s.insert(temp[i]);
                    }
                }

            }
            cnt++;
        }

    }
    if(cnt>1) {
        printf("Error: %d components
",cnt);
    } else {//从第一遍dfs的根结点中任选一个，再次dfs（可以画图模拟一下）
        fill(vis,vis+maxn,0);
        maxdep=0;
        temp.clear();
        dfs(s1,1);
        for(int i=0; i<temp.size(); i++) {
            s.insert(temp[i]);
        }
        for(set<int>::iterator it=s.begin(); it!=s.end(); it++) {
            printf("%d
",*it);
        }
    }

    return 0;
}