three arrays

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题目链接

题意

打乱两个数列，对应位置异或产生一个新数列，使它字典序最小。

思路

建两棵字典树，然后尽量走他们相同的边，这样异或出的数字尽量小，排一下序，让最小的在前面。

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#include<bits/stdc++.h>
using namespace std;
#define ll long long
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50
const int N = 1e5 + 5;
const int maxn = N * 2 * 25;
int tot;
int ch[maxn][2];
int sum[maxn];
int root1, root2;

int newnode()
{
    tot++;
    for (int i = 0; i < 2; i++)
        ch[tot][i] = 0;
    sum[tot] = 0;
    return tot;
}
void insert(int root, int x)
{
    int cur = root;
    for (int p = 29; p >= 0; p--)
    {
        int t = (x >> p) & 1;
        int u = ch[cur][t];
        if (!u)
        {
            ch[cur][t] = newnode();
            u = ch[cur][t];
        }
        sum[cur]++;
        cur = u;
    }
    sum[cur]++;
}
int ans[N];
int cnt;

inline bool exist(int u)
{
    if (!u || sum[u] == 0)
        return false;
    return true;
}

int query(int u, int v, int tmp1, int tmp2)
{
    if (!ch[u][0] && !ch[u][1])
    {
        int s = min(sum[u], sum[v]);
        for (int i = 1; i <= min(sum[u], sum[v]); i++)
            ans[++cnt] = tmp1 ^ tmp2;
        sum[u] -= s;
        sum[v] -= s;
        //dbg(sum[u], sum[v], tmp1, tmp2);
        return s;
    }
    int s = 0;
    for (int i = 0; i < 2; i++)
    {
        if (exist(ch[u][i]) && exist(ch[v][i]))
        {
            //puts("same");
            //dbg(tmp1, tmp2, i);
            int t = query(ch[u][i], ch[v][i], tmp1 << 1 | i, tmp2 << 1 | (i));
            sum[u] -= t;
            sum[v] -= t;
            s += t;
        }
    }
    for (int i = 0; i < 2; i++)
    {
        if (exist(ch[u][i]) && exist(ch[v][!i]))
        {
            //puts("diff");
            //dbg(tmp1, tmp2, i);
            int t = query(ch[u][i], ch[v][!i], tmp1 << 1 | i, tmp2 << 1 | (!i));
            sum[u] -= t;
            sum[v] -= t;
            s += t;
        }
    }
    return s;
}
int a[N], b[N];
void init()
{
    tot = 0;
    root1 = newnode();
    root2 = newnode();
    cnt = 0;
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for (int i = 1; i <= n; i++)
            scanf("%d", &b[i]);
        init();
        for (int i = 1; i <= n; i++)
            insert(root1, a[i]);
        for (int i = 1; i <= n; i++)
            insert(root2, b[i]);
        assert(query(root1, root2, 0, 0) == n);
        sort(ans + 1, ans + n + 1);
        for (int i = 1; i <= n; i++)
            printf("%d%c", ans[i], i == n? '\n': ' ');
    }
    return 0;
}
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