P3038 [USACO11DEC] Grass Planting G

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;
}