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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 |
---|---|---|---|---|---|---|---|---|---|
#583945 | #5418. Color the Tree | Longvu | RE | 0ms | 25192kb | C++14 | 3.9kb | 2024-09-23 00:15:21 | 2024-09-23 00:15:21 |
Judging History
answer
/**
* author: longvu
* created: 09/22/24 22:35:37
**/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define sz(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
const int INF = 1e18;
const int nax = (int)(201001);
const int mod = 1e9 + 7;
template<class X, class Y>
bool maximize(X& x, const Y y) {
if (y > x) {x = y; return true;}
return false;
}
template<class X, class Y>
bool minimize(X& x, const Y y) {
if (y < x) {x = y; return true;}
return false;
}
#define Fi first
#define Se second
const int LOG = 18;
pair<int, int> rmq[nax][LOG + 1];
int euler[nax], h[nax], Log2[nax], fir[nax];
vector<int> adj[nax];
int m = 0;
void dfs(int u, int pa) {
euler[++m] = u;
fir[u] = m;
rmq[m][0] = {h[euler[m]], euler[m]};
for (auto v : adj[u]) {
if (v == pa) {
continue;
}
h[v] = 1 + h[u];
dfs(v, u);
euler[++m] = u;
rmq[m][0] = {h[euler[m]], euler[m]};
}
}
void precal() {
m = 0;
dfs(1, -1);
for (int k = 1; k <= LOG; ++k) {
for (int i = 1; i + (1 << k) - 1 <= m; ++i) {
rmq[i][k] = min(rmq[i][k - 1], rmq[i + (1 << (k - 1))][k - 1]);
}
}
}
int find_lca(int u, int v) {
int l = min(fir[u], fir[v]), r = max(fir[u], fir[v])
, k = Log2[r - l + 1];
return min(rmq[l][k], rmq[r - (1 << k) + 1][k]).Se;
}
int a[nax], dp[nax], rmqp[nax][LOG + 1];
vector<int> memo[nax];
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
for (int i = 2; i < nax; ++i) {
Log2[i] = 1 + Log2[i >> 1];
}
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> a[i];
rmqp[i][0] = a[i];
}
for (int k = 1; k <= LOG; ++k) {
for (int i = 0; i + (1 << k) - 1 < n; ++i) {
rmqp[i][k] = min(rmqp[i][k - 1], rmqp[i + (1 << (k - 1))][k - 1]);
}
}
auto get = [&](int l, int r) {
int k = Log2[r - l + 1];
return min(rmqp[l][k], rmqp[r - (1 << k) + 1][k]);
};
for (int i = 0; i <= n; ++i) {
adj[i].clear();
memo[i] = { -1};
}
for (int i = 1; i < n; ++i) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
precal();
for (int i = 1; i <= n; ++i) {
memo[h[i]].push_back(i);
}
int ans = 0;
for (int hi = 0; hi < n; ++hi) {
sort(1 + all(memo[hi]), [&](int u, int v) {
return fir[u] < fir[v];
});
for (int i = 0; i < sz(memo[hi]); ++i) {
dp[i] = INF;
}
dp[0] = 0;
for (int i = 1; i < sz(memo[hi]); ++i) {
if (sz(memo[hi]) == 1) {
continue;
}
int l = 1, r = i;
auto cal = [&](int z) {
return get(hi - h[find_lca(memo[hi][z], memo[hi][i])], hi) + dp[z - 1];
};
while (r - l >= 1) {
int mid1 = (2 * l + r) / 3, mid2 = (l + 2 * r) / 3;
if (cal(mid1) >= cal(mid2)) {
l = mid1 + 1;
} else {
r = mid2 - 1;
}
}
l--;
for (int j = max(1LL, l - 21); j <= min(l + 21, i); ++j) {
minimize(dp[i], cal(j));
}
assert(dp[i - 1] <= dp[i]);
}
// cout << hi << " " << dp[sz(memo[hi]) - 1] << '\n';
ans += dp[sz(memo[hi]) - 1];
}
cout << ans << '\n';
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 25192kb
input:
3 4 10 15 40 1 1 2 2 3 2 4 5 10 5 1 100 1000 1 2 2 3 2 4 4 5 4 1000 200 10 8 1 2 2 3 3 4
output:
35 17 1218
result:
ok 3 number(s): "35 17 1218"
Test #2:
score: -100
Runtime Error
input:
3000 54 43 44 11 49 17 14 7 30 35 12 34 14 15 8 29 47 30 31 39 17 26 23 26 45 34 50 49 13 35 18 29 15 13 44 47 5 22 20 19 46 30 22 26 13 47 46 43 27 48 48 13 14 30 44 1 2 1 3 2 4 3 5 4 6 2 7 4 8 1 9 3 10 1 11 8 12 2 13 9 14 1 15 1 16 15 17 15 18 7 19 1 20 19 21 13 22 10 23 17 24 9 25 9 26 24 27 8 28...