【BZOJ5293】【BJOI2018】求和

【题目链接】

• 点击打开链接
  

【思路要点】

• 预处理\(i^k\)的前缀和，询问时查询询问点对的LCA，即可将问题转化为\(i^k\)的区间和。
• 时间复杂度\(O(NK+NLogN+QLogN)\)。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
const int MAXLOG = 20;
const int MAXK = 55;
const int P = 998244353;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
vector <int> a[MAXN];
int n, m, val[MAXK][MAXN];
int father[MAXN][MAXK], depth[MAXN];
void work(int pos, int fa, int dep) {
	depth[pos] = dep;
	father[pos][0] = fa;
	for (int i = 1; i < MAXLOG; i++)
		father[pos][i] = father[father[pos][i - 1]][i - 1];
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i] != fa) work(a[pos][i], pos, dep + 1);
}
int lca(int x, int y) {
	if (depth[x] < depth[y]) swap(x, y);
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (depth[father[x][i]] >= depth[y]) x = father[x][i];
	if (x == y) return x;
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (father[x][i] != father[y][i]) {
			x = father[x][i];
			y = father[y][i];
		}
	return father[x][0];
}
int main() {
	read(n);
	for (int i = 1; i <= n - 1; i++) {
		int x, y;
		read(x), read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	for (int i = 1; i <= n; i++) {
		val[0][i] = 1;
		for (int j = 1; j < MAXK; j++)
			val[j][i] = 1ll * val[j - 1][i] * i % P;
	}
	for (int i = 1; i < MAXK; i++)
	for (int j = 1; j <= n; j++)
		val[i][j] = (val[i][j] + val[i][j - 1]) % P;
	depth[0] = -1;
	work(1, 0, 0);
	read(m);
	for (int i = 1; i <= m; i++) {
		int x, y, k;
		read(x), read(y), read(k);
		int z = lca(x, y);
		int ansx = (val[k][depth[x]] - val[k][depth[z]] + P) % P;
		int ansy = (val[k][depth[y]] - val[k][depth[z]] + P) % P;
		int ans = (ansx + ansy) % P;
		if (depth[z] != 0) {
			int tmp = (val[k][depth[z]] - val[k][depth[z] - 1] + P) % P;
			ans = (ans + tmp) % P;
		}
		writeln(ans);
	}
	return 0;
}