树链剖分
原理
重链上结点的编号是连续的。
模板题
#include <iostream>
#include <cstring>
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
typedef long long LL;
const int N = 100010;
struct edge {
int to, next;
};
edge e[2 * N];
int idx, head[N];
int n, m, r, p;
int temp[N], w[N];
int dep[N], fa[N], size[N];
int heavy_son[N], top[N];
int cnt, dfn[N];
struct node {
int l, r;
int data;
int add;
#define l(x) t[x].l
#define r(x) t[x].r
#define data(x) t[x].data
#define add(x) t[x].add
};
node t[N << 2];
void add_edge (int u, int v) {
e[idx].to = v;
e[idx].next = head[u];
head[u] = idx ++;
}
void dfs1 (int cur, int father) {
size[cur] = 1;
for (int i = head[cur]; i != -1; i = e[i].next) {
int to = e[i].to;
if (to == father) continue;
dep[to] = dep[cur] + 1;
fa[to] = cur;
dfs1(to, cur);
size[cur] += size[to];
if (size[to] > size[heavy_son[cur]])
heavy_son[cur] = to;
}
}
void dfs2 (int cur, int top_node) {
dfn[cur] = ++ cnt;
w[dfn[cur]] = temp[cur];
top[cur] = top_node;
if (heavy_son[cur]) dfs2(heavy_son[cur], top_node);
for (int i = head[cur]; i != -1; i = e[i].next) {
int to = e[i].to;
if (to == fa[cur] || to == heavy_son[cur]) continue;
dfs2(to, to);
}
}
void update (int rt) {
data(rt) = (data(ls) + data(rs)) % p;
}
void build (int rt, int l, int r) {
l(rt) = l, r(rt) = r;
if (l == r) {
data(rt) = w[l] % p;
return;
}
int mid = l + r >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
update(rt);
}
void spread (int rt) {
if (add(rt)) {
data(ls) = (data(ls) + (LL)add(rt) * (r(ls) - l(ls) + 1)) % p;
data(rs) = (data(rs) + (LL)add(rt) * (r(rs) - l(rs) + 1)) % p;
add(ls) = (add(ls) + add(rt)) % p;
add(rs) = (add(rs) + add(rt)) % p;
add(rt) = 0;
}
}
void modify (int rt, int l, int r, int val) {
if (l <= l(rt) && r(rt) <= r) {
data(rt) = (data(rt) + (LL)val * (r(rt) - l(rt) + 1)) % p;
add(rt) = (add(rt) + val) % p;
return;
}
spread(rt);
int mid = l(rt) + r(rt) >> 1;
if (l <= mid) modify(ls, l, r, val);
if (r >= mid + 1) modify(rs, l, r, val);
update(rt);
}
int query (int rt, int l, int r) {
if (l <= l(rt) && r(rt) <= r) return data(rt);
spread(rt);
int mid = l(rt) + r(rt) >> 1;
int res = 0;
if (l <= mid) res = (res + query(ls, l, r)) % p;
if (r >= mid + 1) res = (res + query(rs, l, r)) % p;
return res;
}
void modify_path (int x, int y, int val) {
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
modify(1, dfn[top[x]], dfn[x], val);
x = fa[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
modify(1, dfn[x], dfn[y], val);
}
int query_path (int x, int y) {
int res = 0;
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
res = (res + query(1, dfn[top[x]], dfn[x])) % p;
x = fa[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
res = (res + query(1, dfn[x], dfn[y])) % p;
return res;
}
void modify_subtree (int x, int val) {
modify(1, dfn[x], dfn[x] + size[x] - 1, val);
}
int query_subtree (int x) {
return query(1, dfn[x], dfn[x] + size[x] - 1);
}
int main () {
memset(head, -1, sizeof(head));
scanf("%d%d%d%d", &n, &m, &r, &p);
for (int i = 1; i <= n; ++ i) scanf("%d", &temp[i]);
for (int i = 1, u, v; i <= n - 1; ++ i) {
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
dfs1(r, -1);
dfs2(r, r);
build(1, 1, n);
while (m --) {
int opt;
scanf("%d", &opt);
if (opt == 1) {
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
modify_path(x, y, z);
} else if (opt == 2) {
int x, y;
scanf("%d%d", &x, &y);
printf("%d\n", query_path(x, y));
} else if (opt == 3) {
int x, y;
scanf("%d%d", &x, &y);
modify_subtree(x, y);
} else {
int x;
scanf("%d", &x);
printf("%d\n", query_subtree(x));
}
}
return 0;
}