P3313 [SDOI2014] 旅行

题目

思路

为每个宗教维护一个线段数，查询时，树剖时在对应宗教上查询区间即可。

使用动态开点线段树，每次最多新建 \(\log n\) 个节点，不会 MLE。

代码

#include <bits/stdc++.h>

#define range 1, 100000

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;
}

int dep[N], fa[N], sz[N], son[N];

void dfs1(int u, int f) {
    dep[u] = dep[f] + 1, sz[u] = 1, fa[u] = f;
    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[to] > sz[son[u]]) son[u] = to;
    }
}

int top[N], dfn[N], rk[N], dfn_cnt;

void dfs2(int u, int t) {
    dfn[u] = ++dfn_cnt, rk[dfn_cnt] = u, top[u] = t;
    if (son[u]) dfs2(son[u], t);
    for (int i = head[u]; i; i = e[i].next) {
        int to = e[i].to;
        if (to == son[u] || to == fa[u]) continue;
        dfs2(to, to);
    }
}

struct node {
    int l, r;
    int sum;
    int max;
} tr[N * 31];

int root[N], seg_idx;

void pushup(int u) {
    tr[u].sum = tr[tr[u].l].sum + tr[tr[u].r].sum;
    tr[u].max = max(tr[tr[u].l].max, tr[tr[u].r].max);
}

void modify(int& u, int l, int r, int x, int v) {
    if (!u) u = ++seg_idx;
    if (l == r) {
        tr[u].sum = tr[u].max = v;
        return;
    }
    int mid = l + r >> 1;
    if (x <= mid) modify(tr[u].l, l, mid, x, v);
    else modify(tr[u].r, mid + 1, r, x, v);
    pushup(u);
}

pair<int, int> query(int u, int l, int r, int pl, int pr) {
    if (!u) return {0, 0};
    if (pl <= l && r <= pr) return {tr[u].max, tr[u].sum};
    int mid = l + r >> 1;
    if (pr <= mid) return query(tr[u].l, l, mid, pl, pr);
    else if (pl > mid) return query(tr[u].r, mid + 1, r, pl, pr);
    else {
        auto q1 = query(tr[u].l, l, mid, pl, pr);
        auto q2 = query(tr[u].r, mid + 1, r, pl, pr);
        return {max(q1.first, q2.first), q1.second + q2.second};
    }
}

int n, q;
int w[N], c[N];

pair<int, int> ask(int u, int v) {
    pair<int, int> ans = {0, 0};
    int rt = root[c[u]];
    while (top[u] != top[v]) {
        if (dep[top[u]] < dep[top[v]]) swap(u, v);
        auto q = query(rt, range, dfn[top[u]], dfn[u]);
        ans.first = max(ans.first, q.first);
        ans.second += q.second;
        u = fa[top[u]];
    }
    if (dep[u] > dep[v]) swap(u, v);
    auto q = query(rt, range, dfn[u], dfn[v]);
    ans.first = max(ans.first, q.first);
    ans.second += q.second;
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n >> q;

    for (int i = 1; i <= n; i++) cin >> w[i] >> c[i];

    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);

    for (int i = 1; i <= n; i++) modify(root[c[i]], range, dfn[i], w[i]);

    string opt;
    int x, y;
    while (q--) {
        cin >> opt >> x >> y;
        if (opt == "CC") {
            modify(root[c[x]], range, dfn[x], 0);
            c[x] = y;
            modify(root[c[x]], range, dfn[x], w[x]);
        }
        else if (opt == "CW") {
            w[x] = y;
            modify(root[c[x]], range, dfn[x], w[x]);
        }
        else if (opt == "QS") cout << ask(x, y).second << '\n';
        else cout << ask(x, y).first << '\n';
    }
    return 0;
}