【HDOJ6687】Rikka with Stable Marriage（Trie树，贪心）

题意：给定两个长均为n的序列a和b，要求两两配对，a[i]和b[j]配对的值为a[i]^b[j]，求配对后的值之和的最大值

n<=1e5，a[i]，b[i]<=1e9

思路：和字典序最大的策略是等价的

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
//typedef pair<ll,ll>P;
#define N  200010
#define M  200010
#define fi first
#define se second
#define MP make_pair
#define pb push_back
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=1e9+7,inv2=(MOD+1)/2;
      double eps=1e-6;
      ll INF=1e15;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

int t[N*32][2],s[N*32][2],a[N],b[N],m,cnt;
ll ans;

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

void update(int x,int y,int op)
{
    int u=1;
    per(i,30,0)
    {
        int now=(x>>i)&1;
        if(!t[u][now]) t[u][now]=++cnt;
        u=t[u][now];
        s[u][op]+=y;
    }
}

int query(int x,int op)
{
    int u=1,res=0;
    per(i,30,0)
    {
        int now=(x>>i)&1;
        if(t[u][now^1]&&s[t[u][now^1]][op])
        {
            if(now^1) res+=1<<i;
            u=t[u][now^1];
        }
         else
         {
             if(now) res+=1<<i;
             u=t[u][now];
         }

    }
    return res;
}

int find(int op)
{
    int u=1,res=0;
    per(i,30,0)
    {
        if(t[u][0]&&s[t[u][0]][op]) u=t[u][0];
         else
         {
             res+=1<<i;
             u=t[u][1];
         }
    }
    return res;
}

int dfs(int x,int op,int pre)
{
    while(1)
    {
        int y=query(x,op^1);
        if(y==pre)
        {
            m++;
            ans+=(x^y);
            update(x,-1,op);
            update(y,-1,op^1);
            return 1;
        }
        if(dfs(y,op^1,x)) return 0;
    }
}

void solve()
{
    int n=read();
    cnt=1;
    rep(i,1,n)
    {
        a[i]=read();
        update(a[i],1,0);
    }
    rep(i,1,n)
    {
        b[i]=read();
        update(b[i],1,1);
    }
    ans=0;
    m=0;
    while(m<n)
    {
        int x=find(0);
        dfs(x,0,-1);
    }
    printf("%I64d
",ans);
    rep(i,1,cnt)
     rep(j,0,1) t[i][j]=s[i][j]=0;
}

int main()
{
    //freopen("1.in","r",stdin);
    int cas=read();
    while(cas--) solve();
    return 0;
}