2021牛客多校6 K Starch Cat

题目描述

https://ac.nowcoder.com/acm/contest/11257/K

简要题意:给一棵大小为 $n$ 的树,每个点有点权,有 $m$ 次询问,每次询问 $(u,v)$ 的路径上不能选相邻的点的最大点权和

$n\le 5\times 10^5,m\le 10^7$

Solution

注意到 $m$ 有 $10^7$,所以我们肯定不能使用查询 $O(\log n)$ 的算法

另外注意到这个东西可以用 $dp[u][0/1]$ 来维护,这个东西可以直接矩乘维护

我们考虑猫树在点分树上的拓展,对于点分树上的每个子树,我们都预处理子树中的每个点到根的矩阵

为了方便这个矩阵是不包括分支中心的,然后在查询的时候可以 $O(1)rmq$ 求 $lca$ 做到 $O(1)$ 合并信息

时间复杂度 $O(n\log n+m)$

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#include <iostream>
#include <algorithm>
#define maxn 500010
#define ll long long
#define INF 1000000000000000000
using namespace std;

const int p = 998244353;

int n, m, seed, a[maxn];

struct Rand{
unsigned int n,seed;
Rand(unsigned int n,unsigned int seed)
:n(n),seed(seed){}
int get(long long lastans){
seed ^= seed << 13;
seed ^= seed >> 17;
seed ^= seed << 5;
return (seed^lastans)%n+1;
}
};

struct Matrix {
const static int n = 2;
ll a[2][2];

Matrix() { fill(a[0], a[0] + 4, -INF); }

void setone() { for (int i = 0; i < n; ++i) a[i][i] = 0; }

void set(ll x, ll y, ll z, ll w) {
a[0][0] = x; a[0][1] = y;
a[1][0] = z; a[1][1] = w;
}

friend Matrix operator * (const Matrix &u, const Matrix &v) {
Matrix w;
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j) w.a[i][j] = max(u.a[i][1] + v.a[1][j], u.a[i][0] + v.a[0][j]);
return w;
}
} emptyM;

struct Edge {
int to, next;
} e[maxn * 2]; int c1, head[maxn];
inline void add_edge(int u, int v) {
e[c1].to = v; e[c1].next = head[u]; head[u] = c1++;
}

int R[maxn]; bool vis[maxn];
struct Calc_sz {
int f[maxn], sz[maxn];
int sum, rt, maxdp;

void init() { f[rt = 0] = 1e9; }

void dfs_sz(int u, int fa) {
sz[u] = 1;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to; if (v == fa || vis[v]) continue;
dfs_sz(v, u); sz[u] += sz[v];
}
}

void dfs(int u, int fa) {
f[u] = 0;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to; if (v == fa || vis[v]) continue;
dfs(v, u); f[u] = max(f[u], sz[v]);
} f[u] = max(f[u], sum - sz[u]);
if (f[u] < f[rt]) rt = u;
}

inline int get_rt(int u) {
rt = maxdp = 0; dfs_sz(u, 0); sum = sz[u];
dfs(u, 0); return rt;
}
} _;

Matrix f[maxn][21];
void dfs(int u, int fa, int d) {
f[u][d].set(0, 0, a[u], -INF); f[u][d] = f[fa][d] * f[u][d];
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to; if (v == fa || vis[v]) continue;
dfs(v, u, d);
}
}

int id[2 * maxn], in[maxn], dep[maxn];
void divide(int u, int d) {
static int cnt = 0;
id[++cnt] = u; in[u] = cnt; vis[u] = 1; dep[u] = d;
f[u][d].set(0, -INF, -INF, 0);
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to; if (vis[v]) continue;
dfs(v, u, d); divide(_.get_rt(v), d + 1); id[++cnt] = u;
}
f[u][d].set(0, 0, a[u], -INF);
}

inline int st_min(int l, int r) { return in[l] < in[r] ? l : r; }

int st[maxn * 2][21], Log[maxn * 2];
void init_st(int n) { Log[0] = -1;
for (int i = 1; i <= n; ++i) Log[i] = Log[i >> 1] + 1;
for (int i = 1; i <= n; ++i) st[i][0] = id[i];
for (int j = 1; j <= 20; ++j)
for (int i = 1; i + (1 << j) - 1 <= n; ++i)
st[i][j] = st_min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
}

inline int get_lca(int l, int r) {
l = in[l]; r = in[r]; if (l > r) swap(l, r);
int k = Log[r - l + 1];
return st_min(st[l][k], st[r - (1 << k) + 1][k]);
}

ll query(int x, int y) {
int lca = get_lca(x, y), d = dep[lca];
if (dep[x] > dep[y]) swap(x, y);
if (lca == x) return max(max(f[y][d].a[0][0], f[y][d].a[1][0]), a[x] + f[y][d].a[0][0]);
return max(max(f[x][d].a[0][0], f[x][d].a[1][0]) + max(f[y][d].a[0][0], f[y][d].a[1][0]),
a[lca] + f[x][d].a[0][0] + f[y][d].a[0][0]);
}

int main() { fill(head, head + maxn, -1);
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n >> m >> seed;
for (int i = 1; i <= n; ++i) cin >> a[i];
for (int i = 2, x; i <= n; ++i) cin >> x, add_edge(x, i), add_edge(i, x);
_.init(); divide(_.get_rt(1), 1); init_st(2 * n - 1);
ll lans = 0, ans = 0; Rand rand(n, seed);
for (int i = 1; i <= m; ++i) {
int u = rand.get(lans);
int v = rand.get(lans);
int x = rand.get(lans);
lans = query(u, v);
ans = (ans + lans % p * x) % p;
} cout << ans << "\n";
return 0;
}