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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#179320 | #6885. Simple Tree Problem | PPP# | TL | 0ms | 0kb | C++17 | 3.8kb | 2023-09-14 20:31:53 | 2023-09-14 20:31:53 |
answer
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif
//#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int N = 1000500, LOG = 20, mod = 998244353;
int n, a[N], b[N], ans[N];
vector<pair<int, pair<int, int> > > g[N];
int sz[N];
int cnt[N];
int t[N << 2][LOG], t_cnt[N][LOG];
void build(int k, int v, int tl, int tr) {
if (tl == tr) {
t_cnt[tl][k] = 0;
t[v][k] = max(cnt[tl] - t_cnt[tl][k], t_cnt[tl][k]);
return;
}
int tm = (tl + tr) >> 1;
build(k, v << 1, tl, tm);
build(k, v << 1 | 1, tm + 1, tr);
t[v][k] = max(t[v << 1][k], t[v << 1 | 1][k]);
}
void upd(int k, int v, int tl, int tr, int p, int x) {
if (tl == tr) {
t_cnt[p][k] += x;
t[v][k] = max(cnt[p] - t_cnt[p][k], t_cnt[p][k]);
return;
}
int tm = (tl + tr) >> 1;
if (p <= tm)
upd(k, v << 1, tl, tm, p, x);
else
upd(k, v << 1 | 1, tm + 1, tr, p, x);
t[v][k] = max(t[v << 1][k], t[v << 1 | 1][k]);
}
int get(int k, int v, int tl, int tr, int x) {
if (t[v][k] < x)
return -1;
if (tl == tr)
return tl;
int tm = (tl + tr) >> 1;
int res = -1;
if (res == -1)
res = get(k, v << 1 | 1, tm + 1, tr, x);
if (res == -1)
res = get(k, v << 1, tl, tm, x);
return res;
}
void dfs1(int v, int p) {
sz[v] = 1;
for (auto it: g[v]) {
int to = it.first;
if (to == p)
continue;
dfs1(to, v);
sz[v] += sz[to];
}
}
void dfs3(int v, int p, int k, int x) {
upd(k, 1, 0, n - 1, a[v], x);
for (auto it: g[v]) {
int to = it.first;
if (to == p)
continue;
dfs3(to, v, k, x);
}
}
void dfs2(int v, int p, int k) {
int u = -1;
for (auto it: g[v]) {
int to = it.first;
if (to == p)
continue;
if (u == -1 || sz[to] > sz[u])
u = to;
}
u = -1;
for (auto it: g[v]) {
int to = it.first;
if (to == p)
continue;
if (to == u) {
dfs2(to, v, k);
{
int res = get(k, 1, 0, n - 1, it.second.first);
if (res == -1)
res = 0;
else
res = b[res];
ans[it.second.second] = res;
}
} else {
dfs2(to, v, k + 1);
{
int res = get(k + 1, 1, 0, n - 1, it.second.first);
if (res == -1)
res = 0;
else
res = b[res];
ans[it.second.second] = res;
}
dfs3(to, v, k + 1, -1);
dfs3(to, v, k, 1);
}
}
upd(k, 1, 0, n - 1, a[v], 1);
}
void solve() {
cin >> n;
for (int i = 0; i < n; i++) {
g[i].clear();
cnt[i] = 0;
}
for (int i = 0; i < n; i++) {
cin >> a[i];
b[i] = a[i];
}
sort(b, b + n);
for (int i = 0; i < n; i++) {
a[i] = lower_bound(b, b + n, a[i]) - b;
cnt[a[i]]++;
}
for (int i = 0; i < n - 1; i++) {
int v, u, w;
cin >> v >> u >> w;
v--, u--;
g[v].push_back({u, {w, i}});
g[u].push_back({v, {w, i}});
}
dfs1(0, 0);
for (int i = 0; i < LOG; i++)
build(i, 1, 0, n - 1);
dfs2(0, 0, 0);
for (int i = 0; i < n - 1; i++)
cout << ans[i] << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
#ifdef DEBUG
freopen("input.txt", "r", stdin);
#endif
int tst;
cin >> tst;
while (tst--)
solve();
return 0;
}
詳細信息
Test #1:
score: 0
Time Limit Exceeded
input:
10000 96 378804736 378804736 101171470 683875564 378804736 997225055 448759149 683875564 683875564 997225055 152015654 83284224 229458933 101171470 229458933 448759149 448759149 152015654 101171470 600214219 378804736 997225055 448759149 152015654 229458933 229458933 83284224 997225055 229458933 600...
output:
997225055 997225055 0 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 997225055 99722505...