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;
}
标签:int,sum2,sum1,mid,20240923,dp,mod
From: https://www.cnblogs.com/libohan/p/18447263