【题解】P6072 『MdOI R1』Path

思路

可以用启发式合并 Trie 啦！

首先考虑到对于合法的方案，必然存在一个结点 $x$ 使得其中一条路径在 $x$ 的子树中而另一条不在，实际上 $x$ 是两条路径中最浅点之一啦！

所以只需要考虑维护出每个子树内的最优路径和子树外的就好啦！

注意到路径异或和等价于端点的树上前缀异或和的异或和，不妨先处理出来好啦！

子树内的可以直接启发式合并 01 Trie 做啦！

子树外的可以考虑一条最优路径 $p \rightarrow q$，不以这条路径为答案的点只存在于根到 $p, q$ 的路径上，另外处理就好啦！

实际上按顺序遍历一遍，把父亲结点除当前结点之外的子树都加入 Trie 就行啦！

时间复杂度是 $O(n \log n \log V)$ 啦！

绫绫 AI 好耶！！！1 冲冲冲冲冲

代码

view code#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 3e4 + 5;
const int t_sz = maxn * 100;

int n, p, q;
int tot, lg_max;
int a[maxn], fa[maxn], tag[maxn], bel[maxn];
int rt[maxn], in[maxn], out[maxn], rk[maxn];
int cnt[t_sz], son[t_sz][2];
vector<int> g[maxn], w[maxn], v[maxn];

int lg2(int x) { return (x == 0 ? -1 : lg2(x >> 1) + 1); }

void insert(int &t, int v)
{
    if (!t) t = ++tot;
    cnt[t]++;
    for (int i = lg_max, x = t; ~i; i--)
    {
        bool c = v >> i & 1;
        if (!son[x][c]) son[x][c] = ++tot;
        x = son[x][c], cnt[x]++;
    }
}

void clear() { memset(cnt + 1, 0, (tot + 1) * sizeof(int)); }

int query(int t, int v)
{
    int res = 0;
    for (int i = lg_max, x = t; ~i; i--)
    {
        bool c = ~v >> i & 1;
        if (cnt[son[x][c]]) x = son[x][c], res |= (1ll << i);
        else x = son[x][c ^ 1];
    }
    return res;
}

void merge(int t, int p, int cur, int dep, int &v)
{
    if (!t) return;
    if (!~dep)
    {
        v = max(v, query(rt[p], cur));
        insert(rt[p], cur);
        return;
    }
    merge(son[t][0], p, cur, dep - 1, v);
    merge(son[t][1], p, cur | (1 << dep), dep - 1, v);
}

void dfs1(int u)
{
    static int now = 0;
    rk[++now] = u;
    in[u] = 0, insert(rt[u], a[u]);
    for (int i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i], d = w[u][i];
        if (v == fa[u]) continue;
        fa[v] = u, a[v] = a[u] ^ d;
        dfs1(v);
        if (cnt[rt[u]] < cnt[rt[v]]) swap(rt[u], rt[v]);
        merge(rt[v], u, 0, lg_max, in[u]);
        in[u] = max(in[u], in[v]);
    }
}

void dfs2(int u)
{
    clear();
    for (int x = u; x; x = fa[x]) tag[x] = 1;
    int cur = 0;
    for (int k = 1, i; k <= n; v[i].clear(), v[bel[i]].push_back(a[i]), k++)
        if (tag[i = rk[k]]) bel[i] = i;
        else bel[i] = bel[fa[i]];
    for (int k = 1, i; k <= n; k++)
        if (tag[i = rk[k]])
        {
            for (int w : v[fa[i]]) cur = max(cur, query(rt[0], w)), insert(rt[0], w);
            out[i] = max(out[i], cur), tag[i] = 0;
        }
}

int main()
{
    scanf("%d", &n);
    for (int i = 1, u, v, d; i <= n - 1; i++)
    {
        scanf("%d%d%d", &u, &v, &d);
        lg_max = max(lg_max, lg2(d));
        g[u].push_back(v), w[u].push_back(d);
        g[v].push_back(u), w[v].push_back(d);
    }
    dfs1(1);
    clear();
    int va = 0, vb = 0;
    for (int i = 1; i <= n; i++)
    {
        int val = query(rt[0], a[i]);
        insert(rt[0], a[i]);
        out[i] = -1;
        if ((va ^ vb) < val) va = a[i], vb = val ^ va, p = i;
    }
    for (int i = 1; i <= n; i++)
        if (a[i] == vb) q = i;
    dfs2(p), dfs2(q);
    for (int i = 1; i <= n; i++)
        if (!~out[i]) out[i] = va ^ vb;
    int ans = 0;
    // printf("debug %d %d\n", p, q);
    // for (int i = 2; i <= n; i++) printf("%d : %d %d\n", i, in[i], out[i]);
    for (int i = 2; i <= n; i++) ans = max(ans, in[i] + out[i]);
    printf("%d\n", ans);
    return 0;
}

/*
2 : 11 15
3 : 7 11
4 : 3 15
5 : 0 15
6 : 0 15
7 : 7 14
8 : 0 15
9 : 0 14

2 : 11 7
3 : 7 11
4 : 3 7
5 : 0 15
6 : 0 15
7 : 7 14
8 : 0 15
9 : 0 14
*/