树上前缀和
题目链接
多次询问树上的一些路径的权值和
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点前缀和
\(s[i]\)代表从根节点到节点\(i\)的点权和
先自顶向下计算出前缀和\(s[i]\)，然后利用前缀和拼凑\((x, y)\)的路径和
\(s[x] + s[y] - s[lca] - s[fa[lca]]\)
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边前缀和
\(s[i]\)代表从根节点到节点\(i\)的边权和
先自顶向下计算出前缀和\(s[i]\)，然后利用前缀和拼凑\((x, y)\)的路径和
\(s[x] + s[y] - 2 * s[lca]\)
题解
关键点在于如何求最近公共祖先（LCA）
用倍增法求LCA，LCA详情看LCA篇
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define ppb pop_back
#define SZ(v) ((int)v.size())
#define pii pair<int, int>
#define int long long
#define all(v) v.begin(), v.end()
#define debug(x) cout << "======" << x << "========" << "\n"
typedef long long ll;
typedef unsigned int u32;
typedef unsigned long long u64;
typedef double db;
using namespace std;
const int N = 1e6+10;
const int mod = 998244353;
int _;int n, m;
vector<int> e[N];
int dep[N], fa[N][30];
int sum[N][55];
int u, v;int ksm(int a, int b) {a %= mod;b %= mod;int ans = 1;while(b) {if(b&1) {ans = (ans * a) % mod;}b >>= 1;a = (a * a) % mod;}return ans;
}void dfs(int u, int father) {dep[u] = dep[father]+1;for(int i = 1; i <= 50; i++) {sum[u][i] = (ksm(dep[u] - 1, i) + sum[father][i]) % mod;}fa[u][0] = father;for(int i = 1; i <= 19; i++) {fa[u][i] = fa[fa[u][i-1]][i-1];}for(int v : e[u]) {if(v == father) continue;dfs(v, u);}
}int lca(int u, int v) {if(dep[u] < dep[v]) {swap(u, v);}for(int i = 19; i >= 0; i--) {if(dep[fa[u][i]] >= dep[v]) {u = fa[u][i];}}if(u == v) return u;for(int i = 19; i >= 0; i--) {if(fa[u][i] != fa[v][i]) {u = fa[u][i];v = fa[v][i];}}return fa[u][0];
}void solve() {cin >> n;for(int i = 1; i <= n - 1; i++) {cin >> u >> v;e[u].pb(v);e[v].pb(u);}dfs(1, 0);int x, y, k;cin >> m;for(int i = 0; i < m; i++) {cin >> x >> y >> k;int p = lca(x, y);// 注意最后模负数cout << ((sum[x][k] + sum[y][k]) % mod - (sum[p][k] + sum[fa[p][0]][k]) % mod + mod) % mod << "\n"; }
}signed main() {ios::sync_with_stdio(false);cin.tie(nullptr);_ = 1;// cin >> _;while(_--) {solve();} return 0;
}
