题意
思路
我们可以使用树链剖分，将每条边的边权下放，将其当作点权处理，每次操作都要忽略 lca 那个点，因为它所对应的点并不在路径上。
代码
#include <bits/stdc++.h>using namespace std;const int N = 100010;struct edge {int to, next;
} e[N * 2];int head[N], idx = 1;void add(int u, int v) {idx++;e[idx].to = v;e[idx].next = head[u];head[u] = idx;
}struct node {int sum, add;
} tr[N << 2];void pushup(int u) {tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}void addtag(int u, int c, int l, int r) {tr[u].add += c;tr[u].sum += c * (r - l + 1);
}void pushdown(int u, int l, int r) {if (tr[u].add) {int mid = l + r >> 1;addtag(u << 1, tr[u].add, l, mid);addtag(u << 1 | 1, tr[u].add, mid + 1, r);tr[u].add = 0;}
}void modify(int u, int l, int r, int pl, int pr, int v) {if (pl <= l && r <= pr) {addtag(u, v, l, r);return;}pushdown(u, l, r);int mid = l + r >> 1;if (pl <= mid) modify(u << 1, l, mid, pl, pr, v);if (pr > mid) modify(u << 1 | 1, mid + 1, r, pl, pr, v);pushup(u);
}int query(int u, int l, int r, int pl, int pr) {if (pl <= l && r <= pr) return tr[u].sum;pushdown(u, l, r);int mid = l + r >> 1, sum = 0;if (pl <= mid) sum += query(u << 1, l, mid, pl, pr);if (pr > mid) sum += query(u << 1 | 1, mid + 1, r, pl, pr);return sum;
}int son[N], sz[N], fa[N], dep[N];void dfs1(int u, int f) {sz[u] = 1, fa[u] = f, dep[u] = dep[f] + 1;for (int i = head[u]; i; i = e[i].next) {int to = e[i].to;if (to == f) continue;dfs1(to, u);sz[u] += sz[to];if (sz[son[u]] < sz[to]) son[u] = to;}
}int top[N], dfn[N], rk[N], t_cnt;
int n, m;void dfs2(int u, int t) {top[u] = t, dfn[u] = ++t_cnt, rk[t_cnt] = u;if (son[u]) dfs2(son[u], t);for (int i = head[u]; i; i = e[i].next) {int to = e[i].to;if (to == fa[u] || to == son[u]) continue;dfs2(to, to);}
}void update(int u, int v) {while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) swap(u, v);modify(1, 1, n, dfn[top[u]], dfn[u], 1);u = fa[top[u]];}if (dep[u] > dep[v]) swap(u, v);if (u == v) return;modify(1, 1, n, dfn[u] + 1, dfn[v], 1);
}int ask(int u, int v) {int sum = 0;while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) swap(u, v);sum += query(1, 1, n, dfn[top[u]], dfn[u]);u = fa[top[u]];}if (dep[u] > dep[v]) swap(u, v);if (u == v) return sum;sum += query(1, 1, n, dfn[u] + 1, dfn[v]);return sum;
}void solve() {cin >> n >> m;for (int i = 1; i < n; i++) {int u, v;cin >> u >> v;add(u, v), add(v, u);}dfs1(1, 0);dfs2(1, 1);char opt;int x, y;while (m--) {cin >> opt >> x >> y;if (opt == 'P') update(x, y);else cout << ask(x, y) << '\n';}
}int main() {ios::sync_with_stdio(false);cin.tie(nullptr);solve();return 0;
}
