[GXOI/GZOI2019]旧词
相关链接（雾：[LNOI2014]LCA
实际上这题就是加了一个幂。。原题爆破比赛
原题:当\(k=1\)时
暴力：求LCA再求深度。。然后观察一下求LCA的方法。。最暴力的方法就是把根节点到节点\(i\)的路径上的点都打上标记，然后由节点\(y\)往上，直到一个有标记的点为止。。
然而这题并不需要求LCA的序号，只需要深度，于是对于每个询问中\([1,x]\)的每个节点\(i\)，把上面打标记换成权值\(+1\)，把扫到一个有标记的点为止变为统计根节点到节点\(y\)的路径上节点的权值和。。
然后我们发现每个询问中中的节点可以一起做，于是这就变成了树链剖分的模版。。
当\(k\not=1\)时
先观察\(k=1\)时，点的权值其实是\((dep_i)^1-(dep_i-1)^1=1\)
于是，当\(k\not=1\)时，点权就是\((dep_i)^k-(dep_i-1)^k=1\)。当然，处理这个的时候要预处理一下。。其他的和\(k=1\)时一样。
#include <bits/stdc++.h>
#define lson (n << 1)
#define rson (n << 1 | 1)
#define next ___________________________________________________________________________________________________
using namespace std;
const int MAXN = 50004,MOD = 998244353;
int n, q, K, f[MAXN], x, y, value[MAXN], size[MAXN], h, head[MAXN], dep[MAXN], v[MAXN], cnt, t[MAXN << 2], k[MAXN << 2], delta[MAXN << 2], id[MAXN], nid[MAXN], top[MAXN], son[MAXN], ans[MAXN];
struct edge {
int to, next;
} g[MAXN << 1];
struct data {
int x, y, id;
} s[MAXN];
inline void addedge(int x, int y) {g[++h].next = head[x],head[x] = h,g[h].to = y;}
inline int qpow(int a, int b) {
int s = 1;
while (b) {
if (b & 1) s = 1LL * s * a % MOD;
a = 1LL * a * a % MOD,b >>= 1;
}
return s;
}
inline void pushdown(int n) {
int x = delta[n];
delta[n] = 0,
t[lson] = (t[lson] + 1LL * x * k[lson]) % MOD,t[rson] = (t[rson] + 1LL * x * k[rson]) % MOD,
delta[lson] = (delta[lson] + x) % MOD,delta[rson] = (delta[rson] + x) % MOD;
}
inline void dfs1(int x) {
int j, maxs = -1;
size[x] = 1;
for (int i = head[x]; i; i = g[i].next) {
j = g[i].to;
if (dep[j]) continue;
dep[j] = dep[x] + 1,
dfs1(j),
size[x] += size[j];
if (size[j] >= maxs) son[x] = j, maxs = size[j];
}
}
inline void dfs2(int x, int ttop) {
top[x] = ttop,id[x] = ++cnt,nid[cnt] = x;
if (!son[x]) return;
dfs2(son[x], ttop);
int j;
for (int i = head[x]; i; i = g[i].next) {
j = g[i].to;
if (j == son[x] || j == f[x]) continue;
dfs2(j, j);
}
}
inline bool cmp(data a, data b) { return a.x < b.x; }
inline int add(int x, int y) {
x += y;
if (x >= MOD) x -= MOD;
return x;
}
inline void build(int n, int l, int r) {
if (l == r) return (void)(k[n] = v[nid[l]],t[n] = delta[n] = 0);
int mid = (l + r) >> 1;
build(lson, l, mid),build(rson, mid + 1, r),
t[n] = add(t[lson], t[rson]),k[n] = add(k[lson], k[rson]);
}
inline int query(int n, int l, int r, int ll, int rr) {
if (ll <= l && r <= rr) return t[n];
int mid = (l + r) >> 1, ans = 0;
if (delta[n]) pushdown(n);
if (ll <= mid) ans = query(lson, l, mid, ll, rr);
if (rr > mid) ans = add(ans, query(rson, mid + 1, r, ll, rr));
return ans;
}
inline void update(int n, int l, int r, int ll, int rr) {
if (ll <= l && r <= rr) return (void)(t[n] = add(t[n], k[n]),delta[n] = add(delta[n], 1));
int mid = (l + r) >> 1;
if (delta[n]) pushdown(n);
if (ll <= mid) update(lson, l, mid, ll, rr);
if (rr > mid) update(rson, mid + 1, r, ll, rr);
t[n] = add(t[lson], t[rson]);
}
inline void change(int x, int y) {
while (top[x] != top[y]) {
if (dep[top[x]] > dep[top[y]]) swap(x, y);
update(1, 1, n, id[top[y]], id[y]),
y = f[top[y]];
}
if (dep[x] > dep[y]) swap(x, y);
update(1, 1, n, id[x], id[y]);
}
int solve(int x, int y) {
int ans = 0;
while (top[x] != top[y]) {
if (dep[top[x]] > dep[top[y]]) swap(x, y);
ans = add(ans, query(1, 1, n, id[top[y]], id[y])),
y = f[top[y]];
}
if (dep[x] > dep[y]) swap(x, y);
ans = add(ans, query(1, 1, n, id[x], id[y]));
return ans;
}
int main() {
scanf("%d%d%d", &n, &q, &K);
for (int i = 2; i <= n; ++i)
scanf("%d", &f[i]),
addedge(f[i], i);
dep[1] = 1,
dfs1(1),dfs2(1, 1);
for (int i = 1; i <= n; ++i) value[i] = qpow(dep[i], K);
for (int i = 1; i <= n; ++i) v[i] = (value[i] - value[f[i]] + MOD) % MOD;
for (int i = 1; i <= q; ++i)
scanf("%d%d", &s[i].x, &s[i].y),
s[i].id = i;
sort(s + 1, s + 1 + q, cmp),
build(1, 1, n);
int l = 1;
for (int i = 1; i <= q; ++i) {
while (l <= s[i].x && l <= n)
change(1, l),
l++;
ans[s[i].id] = solve(1, s[i].y);
}
for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
return 0;
}
