「LNOI2014」LCA

传送门

这里要用到一个巧妙的想法：我们可以把加上 (dep_u) 转化为节点到根路径加和路径查询。

然后题目要求的是一段区间编号的点对某个点求 ( ext{LCA})，可以考虑离线，然后我们把这道题转化成了树上路径加以及查询。

写个树剖就没了。

参考代码：

#include <algorithm>
#include <cstdio>
using namespace std;

const int _ = 1e5 + 5, mod = 201314;

template < class T > void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

int tot, head[_]; struct Edge { int v, nxt; } edge[_ << 1];
void Add_edge(int u, int v) { edge[++tot] = (Edge) { v, head[u] }, head[u] = tot; }

int n, q, ans[_], cnt;
struct node { int x, y, opt, id; } t[_ << 1];
int cmp(node a, node b) { return a.x < b.x; }
int dep[_], siz[_], son[_], dfn[_], top[_], fa[_], sum[_ << 2], tag[_ << 2];

int lc(int p) { return p << 1; }

int rc(int p) { return p << 1 | 1; }

void pushup(int p) { sum[p] = (sum[lc(p)] + sum[rc(p)]) % mod; }

void add(int p, int v, int l, int r) {
    sum[p] = (sum[p] + v * (r - l + 1) % mod) % mod, tag[p] = (tag[p] + v) % mod;
}

void pushdown(int p, int l, int r, int mid) {
    if (tag[p]) add(lc(p), tag[p], l, mid), add(rc(p), tag[p], mid + 1, r), tag[p] = 0;
}

void update(int ql, int qr, int v, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return add(p, v, l, r);
    int mid = (l + r) >> 1;
    pushdown(p, l, r, mid);
    if (ql <= mid) update(ql, qr, v, lc(p), l, mid);
    if (qr > mid) update(ql, qr, v, rc(p), mid + 1, r);
    pushup(p);
}

int query(int ql, int qr, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return sum[p];
    int mid = (l + r) >> 1, res = 0;
    pushdown(p, l, r, mid);
    if (ql <= mid) res = (res + query(ql, qr, lc(p), l, mid)) % mod;
    if (qr > mid) res = (res + query(ql, qr, rc(p), mid + 1, r)) % mod;
    return res;
}

void dfs(int u, int f) {
    dep[u] = dep[f] + 1, siz[u] = 1, fa[u] = f;
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].v; if (v == f) continue ;
        dfs(v, u), siz[u] += siz[v];
        if (siz[son[u]] < siz[v]) son[u] = v;
    }
}

void dfs(int u, int f, int topf) {
    top[u] = topf, dfn[u] = ++dfn[0];
    if (son[u]) dfs(son[u], u, topf);
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].v; if (v == f || v == son[u]) continue ;
        dfs(v, u, v);
    }
}

void Update(int x) {
    int fx = top[x];
    while (fx != 1)
        update(dfn[fx], dfn[x], 1), x = fa[fx], fx = top[x];
    update(1, dfn[x], 1);
}

int Query(int x) {
    int fx = top[x], res = 0;
    while (fx != 1)
        res = (res + query(dfn[fx], dfn[x])) % mod, x = fa[fx], fx = top[x];
    return (res + query(1, dfn[x])) % mod;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin), freopen("cpp.out", "w", stdout);
#endif
    read(n), read(q);
    for (int x, i = 2; i <= n; ++i) read(x), ++x, Add_edge(x, i), Add_edge(i, x);
    dfs(1, 0), dfs(1, 0, 1);
    for (int l, r, x, i = 1; i <= q; ++i) {
        read(l), ++l, read(r), ++r, read(x), ++x;
        t[++cnt] = (node) { r, x, 1, i };
        t[++cnt] = (node) { l - 1, x, -1, i };
    }
    sort(t + 1, t + cnt + 1, cmp);
    int p = 0;
    for (int i = 1; i <= cnt; ++i) {
        while (p < t[i].x) Update(++p);
        ans[t[i].id] = (ans[t[i].id] + t[i].opt * Query(t[i].y) + mod) % mod;
    }
    for (int i = 1; i <= q; ++i) printf("%d
", ans[i]);
    return 0;
}