【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+,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e15;
       int dx[]={-,,,};
       int dy[]={,,-,};

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

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

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

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

     }
     return res;
 }

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

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

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

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