【树结构】树链剖分简单分析

【树结构】树链剖分

当我们需要在一棵树上完成某些区间操作，而且要求复杂度严格保持在 lg 级别，那么树链剖分往往是不错的选择。

所谓树链剖分，就是把树分割成链，把每条链放到线段树或其他数据结构里面维护。显然，只要我们保证每次区间操作涉及的链的个数为 O(lgn) ，就可以保证总查询或修改复杂度为O(lg2n)。一种常用的分割方式是 “轻重剖分” ，相关资料网上可以查询到。

对于一个区间查询如“对a, b最短路径上的所有节点权值求和”，只需要用倍增处理出c = LCA(a, b)转化为对于一个节点和他祖先节点的区间求和。之后只需要不断检查c是否在当前区间内。如果在直接调用数据结构求和，如果不在则求这一区间的总和，并将节点向上推，直到depth[a] < depth[c]。

以zjoi的树的统计一题为例给出代码：

#include <bits/stdc++.h>using namespace std;const int maxm = 100005, maxn = 100005;int zkw[maxn * 4], sum[maxn * 4], N = 131072, t = 0;void update(int i, int k){    i += N - 1; zkw[i] = sum[i] = k;    for (i >>= 1; i; i >>= 1) {        zkw[i] = max(zkw[i << 1], zkw[(i << 1) + 1]);        sum[i] = sum[i << 1] + sum[(i << 1) + 1];    }}pair<int, int> query(int i, int j){    int ans = 0, maxi = INT_MIN, p, q;    for (p = i + N - 1, q = j + N - 1; p < q; p >>= 1, q >>= 1) {        if (p & 1) { ans += sum[p], maxi = max(maxi, zkw[p]); p++; }        if (!(q & 1)) { ans += sum[q], maxi = max(maxi, zkw[q]); q--; }    }    if (p == q) ans += sum[p], maxi = max(maxi, zkw[p]);    return make_pair(ans, maxi);}struct node {    int to, next;    node() { to = next = 0; }} edge[2 * maxm];int head[maxn], top = 0;int dat[maxn], siz[maxn], id[maxn], ind[maxn], hev[maxn], dep[maxn];int father[maxn][35];int n, m;void push(int i, int j) { edge[++top].to = j; edge[top].next = head[i]; head[i] = top; }int dfs1(int i){    siz[i] = 1;    for (int k = head[i]; k; k = edge[k].next) {        if (!dep[edge[k].to]) {            dep[edge[k].to] = dep[i] + 1; father[edge[k].to][0] = i;            siz[i] += dfs1(edge[k].to);        }    }    return siz[i];}void dfs2(int i, int from){    ind[i] = from; update(++t, dat[i]); id[i] = t;    if (!head[i]) return;    hev[i] = 0;    for (int k = head[i]; k; k = edge[k].next) {        if (dep[edge[k].to] > dep[i] && siz[edge[k].to] > siz[hev[i]])            hev[i] = edge[k].to;    }    if (!hev[i]) {return;}    dfs2(hev[i], from);    for (int k = head[i]; k; k = edge[k].next)        if (dep[edge[k].to] > dep[i] && edge[k].to != hev[i])            dfs2(edge[k].to, edge[k].to);}void travel(int, int);void init(){    dep[1] = 1;    memset(father, 0, sizeof father);    dfs1(1);    dfs2(1, 1);    for (int j = 1; j <= 20; j++)        for (int i = 1; i <= n; i++)            father[i][j] = father[father[i][j-1]][j-1];}inline int lowbit(int i) { return i&(-i); }int lca(int a, int b){    if (dep[a] < dep[b]) swap(a, b);    int dd = dep[a] - dep[b];    while (dd) { a = father[a][(int)(log2(lowbit(dd)))]; dd -= lowbit(dd); }    if (a == b) return a;    for (int i = 20; i >= 0; i--)        if (father[a][i] != father[b][i])            a = father[a][i], b = father[b][i];    return father[a][0];}int query_sum(int i, int j) // j is anc of i{    if (dep[i] < dep[j]) return 0;    if (dep[ind[i]] <= dep[j])        return query(id[j], id[i]).first;    return query_sum(father[ind[i]][0], j) + query(id[ind[i]], id[i]).first;}int query_max(int i, int j){    if (dep[i] < dep[j]) return INT_MIN;    if (dep[ind[i]] <= dep[j])        return query(id[j], id[i]).second;    return max(query_max(father[ind[i]][0], j), query(id[ind[i]], id[i]).second);}inline void change(int i, int j) { update(id[i], j); }inline int read() { int a; scanf("%d", &a); return a; }int main(){    memset(dep, 0, sizeof dep);    memset(head, 0, sizeof head);    memset(hev, 0, sizeof hev);    memset(sum, 0, sizeof sum);    memset(zkw, -127/3, sizeof zkw);    n = read();    for (int i = 1; i < n; i++) {        int a, b; a = read(); b = read();        push(a, b);        push(b, a);    }    for (int i = 1; i <= n; i++)        dat[i] = read();    init();    m = read();    char str[10]; int a, b, c;    for (int i = 1; i <= m; i++) {        scanf("%s", str);        a = read(); b = read();        if (strcmp(str, "CHANGE") == 0) change(a, b);        else if (strcmp(str, "QSUM") == 0) {            c = lca(a, b);            printf("%d\n", query_sum(a, c)+query_sum(b, c)-query_sum(c, c));        }        else {            c = lca(a, b);            printf("%d\n", max(query_max(a, c), query_max(b, c)));        }    }    return 0;}