【Wannafly挑战赛24E】旅行

【Wannafly挑战赛24E】旅行

题面

牛客

题解

首先有一个非常显然的\(dp\)：我们直接把\(s\rightarrow t\)的路径抠出来然后设\(f_{i,j}\)表示到第\(i\)个点，目前余数为\(j\)的方案数。

但是这样子复杂度显然是不对的，我们想办法快速合并对于某个点\(u\)，\(s\rightarrow u\)，\(t\rightarrow u\)的答案。

一般这个点\(u\)都是\(lca(s,t)\)但是我们这道题有一个特别神仙的思路就是将这个点\(u\)设为\(s,t\)在点分树上的\(lca\)。

我们对一组询问，找到\(s,t\)在点分树上的\(lca\)点\(u\)并将询问离线，在点分治到\(u\)时处理询问，那么我们对于\(u\)的询问，\(u\)到所在的这个分治树中间所有点的\(dp\)值我们都是可以求出来的，这样子我们就可以合并答案了。

最后的复杂度是点分治算路径的复杂度和回答询问复杂度，为\(O(kn\log n+qk)\)。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int Mod = 998244353;
const int MAX_N = 2e5 + 5;
struct Graph { int to, next; } e[MAX_N << 1];
int fir[MAX_N], e_cnt;
void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}, fir[u] = e_cnt++; }
int N, K, Q, a[MAX_N];
int Root, Siz, Rmx, size[MAX_N], dep[MAX_N], fa[MAX_N];
bool used[MAX_N];
void getRoot(int x, int fa) {
	size[x] = 1;
	int mx = 0;
	for (int i = fir[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (v == fa || used[v]) continue;
		getRoot(v, x);
		size[x] += size[v];
		mx = max(mx, size[v]);
	}
	mx = max(Siz - size[x], mx);
	if (mx < Rmx) Rmx = mx, Root = x;
}
struct Query { int x, y, id; } ;
vector<Query> vec[MAX_N];
int f[MAX_N][50], ans[MAX_N];
void dfs(int x, int fa) {
	for (int i = fir[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (v == fa || used[v]) continue;
		for (int j = 0; j < K; j++) f[v][j] = f[x][j];
		for (int j = 0; j < K; j++) (f[v][(j + a[v]) % K] += f[x][j]) %= Mod;
		dfs(v, x);
	}
}
void Div(int x, int op) {
	used[x] = 1;
	if (op) {
		for (int i = 0; i < K; i++) f[x][i] = 0;
		f[x][0] = 1;
		dfs(x, 0);
		for (auto i : vec[x]) {
			int tmp[50];
			for (int j = 0; j < K; j++) tmp[j] = f[i.x][j];
			for (int j = 0; j < K; j++) (tmp[(j + a[x]) % K] += f[i.x][j]) %= Mod;
			for (int j = 0; j < K; j++)
				ans[i.id] = (ans[i.id] + 1ll * tmp[j] * f[i.y][(K - j) % K]) % Mod;
		}
	}
	for (int i = fir[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (used[v]) continue;
		Siz = Rmx = size[v];
		getRoot(v, x);
		dep[Root] = dep[x] + 1;
		fa[Root] = x;
		Div(Root, op);
	}
}
int LCA(int x, int y) {
	while (x != y) {
		if (dep[x] < dep[y]) swap(x, y);
		x = fa[x];
	}
	return x;
} 

int main () {
	clearGraph();
	N = gi(), K = gi();
	for (int i = 1; i < N; i++) {
		int u = gi(), v = gi();
		Add_Edge(u, v), Add_Edge(v, u);
	}
	Siz = Rmx = N;
	getRoot(1, 0);
	Div(Root, 0);
	for (int i = 1; i <= N; i++) a[i] = gi() % K;
	Q = gi();
	for (int i = 1; i <= Q; i++) {
		int x = gi(), y = gi();
		vec[LCA(x, y)].push_back((Query){x, y, i});
	}
	memset(used, 0, sizeof(used));
	Siz = Rmx = N;
	getRoot(1, 0);
	Div(Root, 1);
	for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]);
    return 0;
}