20240923

Bouquet

我们可以设计一个状态 \(dp_i\) 表示前 \(i\) 朵花内最多可以选多少朵花,如果第 \(j\) 朵花和第 \(i\) 多花不冲突,要满足以下条件

\[r_j < i 且 l_i > i \]

那么我们可以在 \(r_j\) 时再让 \(j\) 的转移合法,那么只用 \(1 \le j \le r_i\) 那么带修的区间查询是什么数据结构?线段树

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 2e5 + 5;

int n, tr[N * 4], l[N], r[N], dp[N];

vector<pair<int, int>> v[N];

int query(int i, int l, int r, int x, int y) {
  if (x > y) {
    return 0;
  }
  if (l > y || r < x) {
    return 0;
  }
  if (l >= x && r <= y) {
    return tr[i];
  }
  int mid = (l + r) >> 1;
  return max(query(i * 2, l, mid, x, y), query(i * 2 + 1, mid + 1, r, x, y));
}

void change(int i, int l, int r, int p, int x) {
  if (l == r) {
    tr[i] = x;
    return ;
  }
  int mid = (l + r) >> 1;
  if (mid >= p) {
    change(i * 2, l, mid, p, x);
  }
  else change(i * 2 + 1, mid + 1, r, p, x);
  tr[i] = max(tr[i * 2], tr[i * 2 + 1]);
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    l[i] = max(0ll, i - l[i]);
    r[i] = min(n, i + r[i]);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (auto cur : v[i]) {
      change(1, 1, n, cur.first, cur.second);
    }
    dp[i] = query(1, 1, n, 1, l[i] - 1) + 1;
    ans = max(ans, dp[i]);
    v[r[i] + 1].push_back({i, dp[i]});
  }
  cout << ans;
  return 0;
}

龙门对决

我也不会,我太菜了,我还没听懂

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 5e5 + 5, mod = 998244353;

int n, dp[N][2], ans;

vector<int> g[N];

void dfs(int u, int f) {
  dp[u][1] = 1;
  vector<int> sum1, sum2, e;
  sum1.resize(g[u].size() + 5);
  sum2.resize(g[u].size() + 5);
  e.push_back(0);
  for (auto v : g[u]) {
    if (v == f) {
      continue;
    }
    dfs(v, u);
    dp[u][1] *= (dp[v][0] + 1);
    dp[u][1] %= mod;
    e.push_back(v);
  }
  sum1[0] = 1, sum2[e.size()] = 1;
  for (int i = 1; i < e.size(); i++) {
    sum1[i] = sum1[i - 1] * (dp[e[i]][0] + 1);
    sum1[i] %= mod;
  }
  for (int i = e.size() - 1; i > 0; i--) {
    sum2[i] = sum2[i + 1] * (dp[e[i]][0] + 1);
    sum2[i] %= mod;
  }
  int cnt = 0;
  for (auto v : g[u]) {
    if (v == f) {
      continue;
    }
    cnt++;
    dp[u][0] += dp[v][1] * sum1[cnt - 1] % mod * sum2[cnt + 1] % mod;
    dp[u][0] %= mod;
  }
  ans = (ans + dp[u][0]) % mod;
}

signed main() {
  cin >> n;
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}