HDU 4757 tree (persistence dictionary tree)

Link: HDU 4757 tree

Given a tree, each node has a value, and now there is a Q-query, each time you ask about the maximum value of the node value and w or value on the U-V path.

Solution: At first, we thought it was a tree Chain Division. In fact, tree Chain Division is only used to find the LCA (tree Chain Division is not required ).

The resumable dictionary tree copies a new node for all the modified nodes without modifying the original node.

To obtain the status before a change. Consistent with persistent operations, the public data before and after modification is retained.

Create a 01 dictionary tree for the weights of all nodes on the given tree, and each node stores a persistent dictionary tree, indicating that the path from the root node to the node

Dictionary tree formed by path nodes. The process of building each node is obtained by modifying its father's day.

During query, the maximum value or value is determined based on the situation of the three dictionary trees U, V, and LCA (u, v). Note that the node LCA (u, v) must be calculated separately.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;int N, Q, E, V[maxn], first[maxn], jump[maxn * 2], link[maxn * 2];int id, idx[maxn], top[maxn], far[maxn], son[maxn], dep[maxn], cnt[maxn];inline void add_Edge (int u, int v) {    link[E] = v;    jump[E] = first[u];    first[u] = E++;}inline void dfs (int u, int pre, int d) {    far[u] = pre;    son[u] = 0;    dep[u] = d;    cnt[u] = 1;    for (int i = first[u]; i + 1; i = jump[i]) {        int v = link[i];        if (v == pre)             continue;        dfs(v, u, d + 1);        cnt[u] += cnt[v];        if (cnt[son[u]] < cnt[v])            son[u] = v;    }}inline void dfs (int u, int rot) {    idx[u] = ++id;    top[u] = rot;    if(son[u])        dfs(son[u], rot);    for (int i = first[u]; i + 1; i = jump[i]) {        int v = link[i];        if (v == far[u] || v == son[u])             continue;        dfs(v, v);    }}inline int LCA (int u, int v) {    int p = top[u], q = top[v];    while (p != q) {        if (dep[p] < dep[q]) {            swap(p, q);            swap(u, v);        }        u = far[p];        p = top[u];    }    return dep[u] > dep[v] ? v : u;}void init() {    E = id = 0;    memset(first, -1, sizeof(first));    for (int i = 1; i <= N; i++)        scanf("%d", &V[i]);    int u, v;    for (int i = 1; i < N; i++) {        scanf("%d%d", &u, &v);        add_Edge(u, v);        add_Edge(v, u);    }    dfs(1, 0, 0);    dfs(1, 1);}struct node {    int g[2], c;}nd[maxn * 20];int sz, root[maxn];int insert (int r, int w) {    int ret, x;    ret = x = sz++;    nd[x] = nd[r];    for (int i = 15; i >= 0; i--) {        int v = (w>>i)&1;        int t = sz++;        nd[t] = nd[nd[x].g[v]];        nd[t].c++;        nd[x].g[v] = t;        x = t;    }    return ret;}void dfs(int u) {    root[u] = insert(root[far[u]], V[u]);    for (int i = first[u]; i + 1; i = jump[i]) {        int v = link[i];        if (v == far[u])             continue;        dfs(v);    }}void Tire_init() {    sz = 1;    root[0] = nd[0].c = 0;    memset(nd[0].g, 0, sizeof(nd[0].g));    dfs(1);}int query(int x, int y, int z, int w) {    int ans = V[z] ^ w, ret = 0;    z = root[z];    for (int i = 15; i >= 0; i--) {        int v = ((w>>i)&1) ^ 1;        int cnt = nd[nd[x].g[v]].c + nd[nd[y].g[v]].c - 2 * nd[nd[z].g[v]].c;        if (cnt)            ret |= (1<<i);        else            v = v^1;        x = nd[x].g[v], y = nd[y].g[v], z = nd[z].g[v];    }    return max(ans, ret);}int main () {    while (scanf("%d%d", &N, &Q) == 2) {        init();        Tire_init();        int u, v, w;        while (Q--) {            scanf("%d%d%d", &u, &v, &w);            printf("%d\n", query(root[u], root[v], LCA(u, v), w));        }    }    return 0;}

HDU 4757 tree (persistence dictionary tree)