This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define foru(i, l, r) for(int i = l; i <= r; i++)
#define ford(i, r, l) for(int i = r; i >= l; i--)
#define __TIME (1.0 * clock() / CLOCKS_PER_SEC)
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<ii, ii> iiii;
const int N = 2e5 + 5;
const int oo = 1e9, mod = 1e9 + 7;
int n, m, k, color[N], ptr;
string s;
vector<int> adj[N];
int dp[N], depth[N];
void dfs_len(int u, int p = 0) {
dp[u] = depth[u] = depth[p] + 1;
if (depth[ptr] < depth[u]) ptr = u;
for (int v : adj[u]) if (v != p) {
dfs_len(v, u);
dp[u] = max(dp[u], dp[v]);
}
}
int cnt[N], cnt_unique, ans[N];
vector<int> A;
void add(int x) {
if (++cnt[color[x]] == 1) ++cnt_unique;
A.push_back(x);
}
void del() {
int x = A.back(); A.pop_back();
if (--cnt[color[x]] == 0) --cnt_unique;
}
void dfs(int u, int p = 0) {
if (p) add(p);
vector<ii> vec;
for (int v : adj[u]) if (v != p) vec.emplace_back(dp[v], v);
sort(vec.begin(), vec.end(), greater<ii>());
int mx1 = vec.size() > 0 ? vec[0].fi : 0;
int mx2 = vec.size() > 1 ? vec[1].fi : 0;
for (ii v : vec) {
int mx = mx1 + mx2 - max(v.fi, mx2);
while (!A.empty() && mx - depth[u] >= depth[u] - depth[A.back()]) del();
dfs(v.se, u);
}
while (!A.empty() && dp[u] - depth[u] >= depth[u] - depth[A.back()]) del();
ans[u] = max(ans[u], cnt_unique);
}
void process() {
cin >> n >> m;
foru(i,1,n-1) {
int u, v; cin >> u >> v;
adj[u].pb(v); adj[v].pb(u);
}
foru(i,1,n) cin >> color[i];
dfs_len(ptr=1);
int i;
foru(k,0,1) {
dfs_len(i=ptr);
dfs(i);
}
foru(i,1,n) cout << ans[i] << ' ';
return;
}
signed main() {
cin.tie(0)->sync_with_stdio(false);
//freopen(".inp", "r", stdin);
//freopen(".out", "w", stdout);
process();
cerr << "Time elapsed: " << __TIME << " s.\n";
return 0;
}
/*
Xét các trường hợp đặc biệt
Kiểm tra lại input/output
Cố gắng trâu
Lật ngược bài toán
Keep calm and get VOI
Flow:
*/
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