PAT A1021 Deepest Root(25)[最深的根，树]

PAT | AcWing

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 (≤104) 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

给出n个结点（1~n）之间的n条边，问是否能构成⼀一棵树，如果不不能构成则输出它有的连
通分量量个数，如果能构成⼀一棵树，输出能构成最深的树的⾼高度时，树的根结点。如果有多个，按照从
⼩小到⼤大输出。

思路1

1. 以无向图的邻接表的方式存储图
2. 使用并查集判断是否在一个集合中，每次添加一条边的时候双向添加，修改并查集。
3. 若只有一个连通分量，依次计算出所有点为根结点的最大深度，将最大的点记录下来

易错：并查集初始化，头结点初始化，边集需要开点数量的两倍 N * 2 遍历邻接表中的所有点 int i = h[u]; i != -1; i = ne[i]

代码1

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
const int N = 10010;
int h[N], e[N * 2], ne[N * 2], idx;
int p[N];

int findfather(int x)
{
    if(x != p[x]) p[x] = findfather(p[x]);
    return p[x];
}

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int dfs(int u, int father)
{
    int depth = 0;
    for(int i = h[u]; i != -1; i = ne[i])
    {
        int t = e[i];
        if(father == t) continue;
        depth = max(depth, dfs(t, u) + 1);
    }
    return depth;
}

int main()
{
    int n, x, y;
    scanf("%d", &n);

    memset(h, -1, sizeof h);
    for(int i = 1; i <= n; i++)
        p[i] = i;

    int k = n;
    for(int i = 0; i < n - 1; i++)
    {
        scanf("%d%d", &x, &y);
        if(findfather(x) != findfather(y))
        {
            k--;
            p[findfather(x)] = findfather(y);
        }
        add(x, y), add(y, x);
    }
    if(k > 1)
        printf("Error: %d components", k);
    else
    {
        vector<int> ans;
        int max_depth = -1;
        for(int i = 1; i <= n; i++)
        {
            int depth = dfs(i, -1);
            if(depth > max_depth)
            {
                max_depth = depth;
                ans.clear();
                ans.push_back(i);
            }
            else if(depth == max_depth)
                ans.push_back(i);
        }
        for(const auto &item : ans)
            printf("%d\n", item);
    }
    return 0;
}