Find the end points of the diameter of all trees

Given a $For$ a tree with$n$ nodes, solve the endpoints of all diameters in this tree.

Solution: the
first time dfsor bfsstarting from any point, usually choose 1 1Point$1$ , find the distance$1$ dot all the furthest points$points$ . They are the end points of the diameter of the tree.
Then from$points$ choose a pointrtas thedfsorbfsstart point, findrtall the points farthest from the point, they are also the end points of the diameter of the tree.
Since the points of the first traversal and the second traversal may be repeated:
such as a3 3Tree with$3$ points:
edge$1-2$
sides$1-3$
second choice$When$ point$2$ is the starting point,3 3Point$3$ will be selected again, so it needs to be de-duplicated.
You can use$SET$ or a direct manual de-emphasis.

Proof:
Prove why you only need $POIn-TS$ optionally one point to run the second pass.
The initial point of the second traversal is$u$ , the farthest point traversed is$v$
all$u$ 's nearest common ancestor is$root$
then$u-v$ must go through$R\; \&\; ltOOT$ 's.

• If $v$在$In\; points$ , it is equivalent to$v$ One of the farthest places you can run is$u$ , so the points will not be missed.
• If $v$不在$points$中， p o i n t s points The farthest other points in$p$$o$$i$$n$$t$$s$ reach must also have$v$ , which is equivalent to running at a different starting point, but these starting points go to$Root$ are all the same distance, and they must pass through the$root$ reach$v$ , so the distance is:$dis(root,u)+dis(root,v)$

Example: the
deepest root

Sample code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

#define sz(x) (int)x.size()

const int N = 20010, M = 20010;
int n;
int h[N], e[M], ne[M], idx;
int p[N];
int ans[N], g;
int dis[N];
void add(int a, int b) {
    
    
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

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

int rt, mx = -1;
void dfs(int u, int fa, int dep) {
    
    
    if(dep > mx) mx = dep, rt = u;
    dis[u] = dep;
    for(int i = h[u]; ~i; i = ne[i]) {
    
    
        int v = e[i];
        if(v == fa) continue;
        dfs(v, u, dep + 1);
    }
}

void dfs2(int u, int fa, int dep) {
    
    
    mx = max(mx, dep);
    dis[u] = dep;
    for(int i = h[u]; ~i; i = ne[i]) {
    
    
        int v = e[i];
        if(v == fa) continue;
        dfs2(v, u, dep + 1);
    }
}

int main()
{
    
    
    scanf("%d", &n);
    memset(h, -1, n + 1 << 2);
    int cnt = n;
    for(int i = 1; i <= n; ++i) p[i] = i;
    for(int i = 1; i < n; ++i) {
    
    
        int a, b; scanf("%d%d", &a, &b);
        if(find(a) != find(b)) {
    
    
            p[find(a)] = p[find(b)];
            --cnt;
        }
        add(a, b);
        add(b, a);
    }
    
    if(cnt != 1) printf("Error: %d components\n", cnt);
    else {
    
    
        dfs(1, -1, 0);
        for(int i = 1; i <= n; ++i)
            if(dis[i] == mx) ans[++g] = i;
            
        mx = -1;
        dfs2(rt, 0, 0);
        
        for(int i = 1; i <= n; ++i) 
            if(dis[i] == mx) ans[++g] = i;
        sort(ans + 1, ans + g + 1);
        n = unique(ans + 1, ans + g + 1) - ans - 1;
        for(int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
    }
    
    return 0;
}