Description
The company $X$ has $n$ employees numbered from $1$ through $n$. Each employee $u$ has a direct boss $p_u$ $(1 \le p_u \le n)$, except for the employee $1$ who has no boss. It is guaranteed, that values $p_i$ form a tree. Employee $u$ is said to be in charge of employee $v$ if $u$ is the direct boss of $v$ or there is an employee $w$ such that $w$ is in charge of $v$ and $u$ is the direct boss of $w$. Also, any employee is considered to be in charge of himself.node
In addition, for each employee $u$ we define it's level $lv(u)$ as follow:c++
- $lv(1)=0$
- $lv(u)=lv(p_u)+1$ for $u \neq 1$
In the near future, there are $q$ possible plans for the company to operate. The $i$-th plan consists of two integers $l_i$ and $r_i$, meaning that all the employees in the range $[l_i,r_i]$, and only they, are involved in this plan. To operate the plan smoothly, there must be a project manager who is an employee in charge of all the involved employees. To be precise, if an employee $u$ is chosen as the project manager for the $i$-th plan then for every employee $v \in [l_i,r_i]$, $u$ must be in charge of $v$. Note, that $u$ is not necessary in the range $[l_i,r_i]$. Also, $u$ is always chosen in such a way that $lv(u)$ is as large as possible (the higher the level is, the lower the salary that the company has to pay the employee).ide
Before any plan is operated, the company has JATC take a look at their plans. After a glance, he tells the company that for every plan, it's possible to reduce the number of the involved employees exactly by one without affecting the plan. Being greedy, the company asks JATC which employee they should kick out of the plan so that the level of the project manager required is as large as possible. JATC has already figured out the answer and challenges you to do the same.ui
Input
The first line contains two integers $n$ and $q$ $(2 \le n \le 100000, 1 \le q \le 100000)$ — the number of employees and the number of plans, respectively.this
The second line contains $n−1$ integers $p_2,p_3,…,p_n(1≤p_i≤n)$ meaning $p_{i}$ is the direct boss of employee $i$.spa
It is guaranteed, that values $p_{i}$ form a directed tree with the root of $1$.code
Each of the following $q$ lines contains two integers $l_i$ and $r_i$ $(1 \le l_i < r_i \le n)$ — the range of the employees, involved in the corresponding plan.orm
Output
Print $q$ lines, each containing two integers — the number of the employee which should be kicked from the corresponding plan and the maximum possible level of the project manager in that case.ip
If there are more than one way to choose that employee, print any of them.ci
Example
Input
11 5 1 1 3 3 3 4 2 7 7 6 4 6 4 8 1 11 9 11 8 11
Output
4 1 8 1 1 0 11 3 8 1
Note
In the example:
In the first query, we can choose whether $4$ or $5$ or $6$ and the project manager will be $3$.
In the second query, if we choose any employee other than the employee $8$, the project manager will be $1$. If we choose $8$, the project manager will be $3$. Since $lv(3)=1 \gt lv(1)=0$, choosing $8$ is the best strategy.
In the third query, no matter how we choose the employee, the project manager will always be $1$.
In the fourth query, if we choose $9$ or $10$ then the project manager will be $3$. If we choose $11$ then the project manager will be $7$. Since $lv(7)=3 \gt lv(3)=1$, we choose $11$ as the answer.
Solution
題意：給一棵樹，$n$個點，$q$次詢問，每次詢問給定一個區間$[l, r]$，要求忽略掉$[l, r]$中的一個點，使得剩下的$r - l $個點的LCA的深度最大，問應該忽略哪一個點，忽略後的最大深度是多少。
首先求一次DFS序，對於任意點$u$，其DFS序記爲$order[u]$。給定區間$[l, r]$，設其中DFS序最大和最小的點分別爲$u$和$v$，則$LCA[l, r]$就是$LCA(u, v)$。咱們能夠簡單證實一下，不妨設$r = LCA(u, v)$，點$x$不屬於以$r$爲根的子樹（記做$SubTree(r)$）當且僅當$order[x]$知足如下兩種狀況中的一種：
- $order[x] \lt order[r]$，即$x$在$r$以前被訪問
- $order[x] > order[i], \forall i \in SubTree(r)$，即$x$在$SubTree(r)$以後才被訪問
顯然，$[l, r]$中的任何一個點都不知足上述兩個條件，因此$[l, r]$中的每一個點都屬於以$r$爲根的子樹，因此它們的LCA就是$r$。
回到咱們的問題，對於每次詢問，給定$[l, r]$，咱們先求出其中DFS序最大、最小的點$u, v$以及它們的LCA $r$。顯然，忽略$u$和$v$以外的節點對並不會改變LCA；若是忽略$u$，那麼新的LCA就是$LCA[l, u-1]$和$LCA[u + 1, r]$的LCA，咱們稱之爲$r_1$；同理，忽略$v$也能夠獲得一個新的LCA，咱們稱之爲$r_2$。選擇$r, r_1, r_2$中深度最大的點，咱們就獲得了答案。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100011;
const int maxp = 18;
vector<int> w[maxn];
int idx, dfsod[maxn], invdfsod[maxn];
int fa[maxn][maxp], dep[maxn];
void bfs(int root) {
queue<int> que;
dep[root] = 0;
fa[root][0] = root;
que.push(root);
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = 1; i < maxp; i++)
fa[u][i] = fa[fa[u][i - 1]][i - 1];
for (int v : w[u]) {
if (v == fa[u][0]) continue;
dep[v] = dep[u] + 1;
fa[v][0] = u;
que.push(v);
}
}
}
int lca(int u, int v) {
if (dep[u] > dep[v])
swap(u, v);
for (int gap = dep[v] - dep[u], i = 0; gap; gap >>= 1, i++) {
if (gap & 1)
v = fa[v][i];
}
if (u == v) return u;
for (int i = maxp - 1; i >= 0; i--) {
if (fa[u][i] == fa[v][i])
continue;
u = fa[u][i], v = fa[v][i];
}
return fa[u][0];
}
void dfs(int u, int pre = -1) {
dfsod[u] = ++idx;
invdfsod[idx] = u;
for (int v : w[u]) {
if (v == pre) continue;
dfs(v, u);
}
}
struct node {
int l, r, mx, mn;
} seg[maxn << 2];
void pushup(int x) {
seg[x].mx = max(seg[x << 1].mx, seg[x << 1 | 1].mx);
seg[x].mn = min(seg[x << 1].mn, seg[x << 1 | 1].mn);
}
void build(int x, int l, int r) {
seg[x].l = l, seg[x].r = r;
if (l == r) {
seg[x].mx = seg[x].mn = dfsod[l];
return;
}
int m = (l + r) >> 1;
build(x << 1, l, m);
build(x << 1 | 1, m + 1, r);
pushup(x);
}
pair<int, int> query(int x, int l, int r) {
int L = seg[x].l, R = seg[x].r;
if (l <= L && r >= R)
return make_pair(seg[x].mn, seg[x].mx);
int m = (L + R) >> 1;
int mx = 0, mn = 1 << 30;
if (l <= m) {
auto v = query(x << 1, l, r);
mn = min(mn, v.first);
mx = max(mx, v.second);
}
if (r > m) {
auto v = query(x << 1 | 1, l, r);
mn = min(mn, v.first);
mx = max(mx, v.second);
}
return make_pair(mn, mx);
}
// 區間[l, r]的LCA
int getlca(int l, int r) {
if (l > r) return -1;
auto x = query(1, l, r);
int u = invdfsod[x.first], v = invdfsod[x.second];
return lca(u, v);
}
// 忽略u後，區間[l, r]的LCA
int getlca(int l, int r, int u) {
int a = getlca(l, u - 1), b = getlca(u + 1, r);
if (a == -1) return b;
if (b == -1) return a;
return lca(a, b);
}
int main() {
int n, q;
scanf("%d%d", &n, &q);
for (int i = 2; i <= n; ++i) {
int x; scanf("%d", &x);
w[x].push_back(i);
w[i].push_back(x);
}
bfs(1);
dfs(1);
build(1, 1, n);
dep[0] = -1;
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
auto x = query(1, l, r);
int u = invdfsod[x.first], v = invdfsod[x.second];
int c = lca(u, v), a = getlca(l, r, u), b = getlca(l, r, v);
int mx = max(dep[c], max(dep[a], dep[b])), y;
if (mx == dep[c]) y = l;
else if (mx == dep[a]) y = u;
else y = v;
printf("%d %d\n", y, mx);
}
return 0;
}