树链剖分——轻重链剖分

2022年01月27日，第十三天

1. 题目链接：P3384 【模板】轻重链剖分/树链剖分

思路：树链剖分直接搞，时间复杂度为 O(nlog2n)O(nlog^2n)O(nlog2n) ，通过模板题，我们可以了解到，树链剖分是用来划分树的路径的一个有效工具，它把任意两点之间的简单路径划分成 O(logn)O(log\ n)O(log n) 段线性区间，当我们得到 dfsdfsdfs 序之后套上线段树时，能够轻松地维护树上的某条链的信息。细节看代码注释。

#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define re register
#define endl '\n'using namespace std;const int MOD = 998244353;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double Pi = acos(-1.0);
const int N = 1e5 + 5, M = N << 1;#define ls (x << 1)
#define rs (x << 1 | 1)int n, m, w[N], rt, P;
int h[N], e[M], ne[M], idx; // 链式前向星
int id[N], nw[N], cnt;
int dep[N]; // 当前结点的深度
int sz[N];  // 当前结点子树的大小
int top[N]; // 当前结点所在重链的最上的结点编号
int fa[N];  // 当前结点的父亲节点
int son[N]; // 当前结点的重儿子
struct Tree {int l, r;ll add, sum;
}tr[N << 2]; // 线段树信息
void add_edge (int u, int v) {e[idx] = v;ne[idx] = h[u];h[u] = idx ++;
}void dfs1 (int u, int p, int depth) {  // 用来处理重儿子dep[u] = depth, fa[u] = p, sz[u] = 1;for (int i = h[u]; ~ i; i = ne[i]) {int j = e[i];if (j == p) continue;dfs1 (j, u, depth + 1);   sz[u] += sz[j];   if (sz[son[u]] < sz[j]) son[u] = j; // 所有儿子中子树最大为重儿子}
}void dfs2 (int u, int t) { // 处理 dfs 序id[u] = ++ cnt, nw[cnt] = w[u], top[u] = t;if (! son[u]) return ;dfs2 (son[u], t);  // 重儿子的 top 是轻儿子传下来的for (int i = h[u]; ~ i; i = ne[i]) {int j = e[i];if (j == fa[u] || j == son[u]) continue;dfs2 (j, j);    // 轻儿子的 top 是自己}
}void pushup (int x) {tr[x].sum = (tr[ls].sum + tr[rs].sum) % P;
}void pushdown (int x) {Tree &p = tr[x], &lson = tr[ls], &rson = tr[rs];if (p.add) {lson.add = (p.add + lson.add) % P;lson.sum = (lson.sum + p.add * (lson.r - lson.l + 1) % P) % P;rson.add = (p.add + rson.add) % P;rson.sum = (rson.sum + p.add * (rson.r - rson.l + 1) % P) % P;p.add = 0;}
}void build (int l, int r, int x = 1) {tr[x] = {l, r, 0, nw[r]};if (l == r) return;int mid = (l + r) >> 1;build (l, mid, ls), build (mid + 1, r, rs);pushup (x);
}void update (int l, int r, int v, int x = 1) {if (l <= tr[x].l && tr[x].r <= r) {tr[x].add = (tr[x].add + v) % P;tr[x].sum = (tr[x].sum + (ll)v * (tr[x].r - tr[x].l + 1) % P) % P;return ;}pushdown (x);int mid = (tr[x].l + tr[x].r) >> 1;if (l <= mid) update (l, r, v, ls);if (r >  mid) update (l, r, v, rs);pushup (x);
}ll query (int l, int r, int x = 1) {if (l <= tr[x].l && tr[x].r <= r) return tr[x].sum;pushdown (x);int mid = (tr[x].l + tr[x].r) >> 1;ll res = 0;if (l <= mid) res += query (l, r, ls);if (r >  mid) res += query (l, r, rs);return res % P;
}void update_path (int x, int y, int v) {while (top[x] != top[y]) {if (dep[top[x]] < dep[top[y]]) swap (x, y);update (id[top[x]], id[x], v);x = fa[top[x]];}if (dep[x] < dep[y]) swap (x, y);update (id[y], id[x], v);
}ll ask_path (int x, int y) {ll res = 0;while (top[x] != top[y]) {if (dep[top[x]] < dep[top[y]]) swap (x, y);res = (res + query (id[top[x]], id[x])) % P;x = fa[top[x]];}if (dep[x] < dep[y]) swap (x, y);return (res += query (id[y], id[x])) % P;
}void update_tree (int x, int v) {update (id[x], id[x] + sz[x] - 1, v);
}ll ask_tree (int x) {return query (id[x], id[x] + sz[x] - 1);
}int main() {ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> n >> m >> rt >> P;for (int i = 1; i <= n; i ++) cin >> w[i];memset (h, -1, sizeof (h));for (int i = 1; i < n; i ++) {int u, v; cin >> u >> v;add_edge (u, v), add_edge (v, u);}dfs1 (rt, -1, 1);    // 处理重儿子dfs2 (rt, rt);      // 处理 dfs 序build (1, n);int op, x, y, z;while (m --) {cin >> op;if (op == 1) {cin >> x >> y >> z;update_path (x, y, z);}if (op == 2) {cin >> x >> y;cout << ask_path (x, y) << endl;}if (op == 3) {cin >> x >> z;update_tree (x, z);}if (op == 4) {cin >> x;cout << ask_tree (x) << endl;}}return 0;
}

2. 题目链接：P2146 [NOI2015] 软件包管理器

思路：树链剖分裸题，时间复杂度 O(nlog2n)O(nlog^2n)O(nlog2n) 。

#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define re register
#define endl '\n'using namespace std;const int MOD = 998244353;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double Pi = acos(-1.0);
const int N = 1e5 + 5;int h[N], e[N], ne[N], idx;
int dep[N], sz[N], top[N], fa[N], son[N];
int id[N], cnt, n, qur;
struct node {int l, r, lazy, sum;
}t[N << 2];void add_edge (int u, int v) {e[idx] = v;ne[idx] = h[u];h[u] = idx ++;
}void dfs1 (int u, int depth) {dep[u] = depth, sz[u] = 1;for (int i = h[u]; ~ i; i = ne[i]) {int v = e[i];dfs1 (v, depth + 1);sz[u] += sz[v];if (sz[son[u]] < sz[v]) son[u] = v;}
}void dfs2 (int u, int T) {id[u] = ++ cnt, top[u] = T;if (! son[u]) return ;dfs2 (son[u], T);for (int i = h[u]; ~ i; i = ne[i]) {int v = e[i];if (v == son[u]) continue;dfs2 (v, v);}
}#define ls (x << 1)
#define rs (x << 1 | 1)void build (int l, int r, int x = 1) {t[x] = {l, r, -1, 0};if (l == r) return ;int mid = (l + r) >> 1;build (l, mid, ls), build (mid + 1, r, rs);
}void pushdown (int x) {node &p = t[x], &lson = t[ls], &rson = t[rs];if (p.lazy != -1) {lson.lazy = rson.lazy = p.lazy;lson.sum = p.lazy * (lson.r - lson.l + 1);rson.sum = p.lazy * (rson.r - rson.l + 1);p.lazy = -1;}
}void pushup (int x) {t[x].sum = t[ls].sum + t[rs].sum;
}void update (int l, int r, int v, int x = 1) {  if (l <= t[x].l && t[x].r <= r) {t[x].lazy = v;t[x].sum = v * (t[x].r - t[x].l + 1);return ;}pushdown (x);int mid = (t[x].l + t[x].r) >> 1;if (l <= mid) update (l, r, v, ls);if (r >  mid) update (l, r, v, rs);pushup (x);
}int query (int l, int r, int x = 1) {if (l <= t[x].l && t[x].r <= r) return t[x].sum;pushdown (x);int mid = (t[x].l + t[x].r) >> 1, res = 0;if (l <= mid) res += query (l, r, ls);if (r >  mid) res += query (l, r, rs);return res;
}void update_path (int v, int x, int y = 1) {while (top[x] != top[y]) {if (dep[top[x]] < dep[top[y]]) swap (x, y);update (id[top[x]], id[x], v);x = fa[top[x]];}if (dep[x] < dep[y]) swap (x, y);update (id[y], id[x], v);
}void update_tree (int x, int v) {update (id[x], id[x] + sz[x] - 1, v);
}int main() {ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> n, fa[1] = 1;memset (h, -1, sizeof (h));for (int i = 2; i <= n; i ++) {int p; cin >> p, p ++;add_edge (p, i), fa[i] = p;}dfs1 (1, 1), dfs2 (1, 1);build (1, n);cin >> qur;while (qur --) {int x; char op[20]; cin >> op >> x, x ++;if (* op == 'i') {int now = t[1].sum;update_path (1, x);cout << t[1].sum - now << endl;}else {int now = t[1].sum;update_tree (x, 0);cout << now - t[1].sum << endl;}}return 0;
}

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树链剖分——轻重链剖分

1. 题目链接：P3384 【模板】轻重链剖分/树链剖分

2. 题目链接：P2146 [NOI2015] 软件包管理器

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