HDU4757： Tree（可持久化Trie）

最新推荐文章于 2022-08-18 13:38:39 发布

原创最新推荐文章于 2022-08-18 13:38:39 发布 · 428 阅读

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本文介绍了一种解决特定树形结构上的路径异或最大值问题的方法，通过使用可持久化 Trie 树来记录从每个节点到根节点的历史信息，并结合经典 01 字典树的做法，有效地解决了给定两个节点间路径上选取一点与给定值进行异或运算的最大值问题。

Zero and One are good friends who always have fun with each other. This time, they decide to do something on a tree which is a kind of graph that there is only one path from node to node. First, Zero will give One an tree and every node in this tree has a value. Then, Zero will ask One a series of queries. Each query contains three parameters: x, y, z which mean that he want to know the maximum value produced by z xor each value on the path from node x to node y (include node x, node y). Unfortunately, One has no idea in this question. So he need you to solve it.

Input  There are several test cases and the cases end with EOF. For each case:

  The first line contains two integers n(1<=n<=10^5) and m(1<=m<=10^5), which are the amount of tree’s nodes and queries, respectively.

  The second line contains n integers a[1..n] and a[i](0<=a[i]<2^{16}) is the value on the ith node.

  The next n–1 lines contains two integers u v, which means there is an connection between u and v.

  The next m lines contains three integers x y z, which are the parameters of Zero’s query.
Output  For each query, output the answer.
Sample Input

Sample Output

3
0

题意：一颗带权树，M个询问，每个询问给出三个值u、v、w，问树上u到v路径中选一个点异或w的最大值是多少？

思路：用可持久化Tri记录每个点到根节点的历史信息，然后就是经典01字典树做法了。

# include <iostream>
# include <cstdio>
# include <algorithm>
# include <vector>
using namespace std;
const int maxn = 1e5+30;
int n, m;
vector<int>g[maxn];
int fa[maxn][20], a[maxn], deep[maxn];
int tree[maxn*35][2], son[maxn*35][2], root[maxn], cnt;
void build(int &x, int y, int val, int pos)
{
    if(pos<0) return;
    x = ++cnt;
    int tmp = !!(1<<pos&val);
    tree[x][tmp] = tree[y][tmp] + 1;
    tree[x][tmp^1] = tree[y][tmp^1];
    son[x][tmp^1] = son[y][tmp^1];
    build(son[x][tmp], son[y][tmp], val, pos-1);
}
int query(int x, int y, int z, int val, int sum, int pos)
{
    if(pos < 0) return sum;
    int tmp = !!(1<<pos&val);
    if(tree[x][tmp^1]+tree[y][tmp^1]-2*tree[z][tmp^1] > 0)
        return query(son[x][tmp^1], son[y][tmp^1], son[z][tmp^1], val, 1<<pos|sum, pos-1);
    else
        return query(son[x][tmp], son[y][tmp], son[z][tmp], val, sum, pos-1);
}
void dfs(int u, int pre)
{
    deep[u] = deep[pre] + 1;
    fa[u][0] = pre;
    for(int i=1; i<20; ++i) fa[u][i] = fa[fa[u][i-1]][i-1];
    build(root[u], root[pre], a[u], 15);
    for(int i=0; i<g[u].size(); ++i)
    {
        int v = g[u][i];
        if(v != pre) dfs(v, u);
    }
}
int lca(int v, int u)
{
    if(deep[v] > deep[u]) swap(u,v);
    for(int i=0; i<20; ++i)
        if(deep[u]-deep[v]>>i&1)
            u = fa[u][i];
    for(int i=19; ~i; --i)
        if(fa[v][i] != fa[u][i])
            v=fa[v][i], u=fa[u][i];
    return v==u?v:fa[v][0];
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        cnt = 0;
        for(int i=1; i<=n; ++i)
        {
            scanf("%d",&a[i]);
            g[i].clear();
        }
        for(int i=1, u, v; i<n; ++i)
        {
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1, 0);
        while(m--)
        {
            int u, v, w;
            scanf("%d%d%d",&u,&v, &w);
            int l = lca(u,v), ans;
            ans = max(w^a[l], query(root[u],root[v],root[l],w,0,15));
            printf("%d\n",ans);
        }
    }
    return 0;
}