逆天周赛题，考虑先缩连通块，然后选一条直径扩张，每次会把周围的不同色点“吸”进来，答案是直径一半上取整。

充分必要性都容易证明吧。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <iostream>
#include <vector>
#include <queue>
#include <cassert>

using namespace std;

#define PII pair<int,int>
#define x first
#define y second
#define For(i, l, r) for(int i = l; i <= r; i ++)
#define Rep(i, l, r) for(int i = l; i >= r; i --)

bool START;

void in(int &x)
{
    char c = getchar(); int op = 1;
    while(c > '9' || c < '0') op *= (c == '-' ? -1 : 1), c = getchar();
    for(x = 0; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0'; x *= op;
}

const int N = 2e5 + 50;

int n, m, f[N], a[N], z[3], fa[N];
vector<int> G[N];
int ans;

void Dfs(int u, int pa)
{
    for(int v : G[u]) if(v != pa)
    {
        Dfs(v, u);
        ans = max(ans, f[u] + f[v] + 1);
        f[u] = max(f[u], f[v] + 1);
    }
}

int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}

PII p[N];

bool ENDPOS = 0;
int main()
{
    in(n);
    For(i, 1, n) in(a[i]);
    For(i, 1, n) fa[i] = i;
    For(i, 1, n - 1)
    {
        int u, v; in(u); in(v); p[i] = {u, v};
        if(a[u] == a[v]) fa[find(u)] = find(v);
    }
    For(i, 1, n - 1)
    {
        int u = p[i].x, v = p[i].y;
        if(find(u) == find(v)) continue;
        u = find(u); v = find(v);
        G[u].push_back(v); G[v].push_back(u);
    }
    int rt = find(1);
    Dfs(rt, 0);
    printf("%d\n", (ans + 1) / 2);
    return 0;
}