【HDOJ6625】three arrays（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]，求字典序最小的配对后的值序列

n<=1e5，a[i],b[i]<2^30

思路：

做法一：orz ckw大佬

 #include<bits/stdc++.h>
 using namespace std;
 const int MAX_N=+5e6;

 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;
 }

 vector<int> ans;
 struct SEG{
     struct Node{
         int son[],sz1,sz2;
     }tree[MAX_N];
     inline int new_node(){
         tree[++top]=(Node){{,},,};
         return top;
     }
     int rt,top;
     inline void up(int x){
         tree[x].sz1=tree[tree[x].son[]].sz1+tree[tree[x].son[]].sz1;
         tree[x].sz2=tree[tree[x].son[]].sz2+tree[tree[x].son[]].sz2;
     }
     void clear(){ top=,rt=new_node(); }
     void insert(int pos,int k1,int k2){
         int x=rt;
         tree[x].sz1+=k1;
         tree[x].sz2+=k2;
         for(int i=;i>=;--i){
             bool t=pos&(<<i);
             if(tree[x].son[t]==)
                 tree[x].son[t]=new_node();
             x=tree[x].son[t];
             tree[x].sz1+=k1;
             tree[x].sz2+=k2;
         }
     }
     int merge(int x,int y){
         if(x==||y==) return x+y;
         tree[x].sz1+=tree[y].sz1;
         tree[x].sz2+=tree[y].sz2;
         tree[x].son[]=merge(tree[x].son[],tree[y].son[]);
         tree[x].son[]=merge(tree[x].son[],tree[y].son[]);
         return x;
     }
     void solve(int x,int key,int p){
 //        printf("{%d %d %d}",x,key,p);
         if(p==){
             int t=min(tree[x].sz1,tree[x].sz2);
 //            printf("[%d]",t);
             for(int i=;i<=t;++i) ans.push_back(key);
             tree[x].sz1-=t;
             tree[x].sz2-=t;
             return;
         }
         if(tree[x].sz1==||tree[x].sz2==) return;
         solve(tree[x].son[],key,p-);
         solve(tree[x].son[],key,p-);
         up(x);
         if(tree[x].sz1==||tree[x].sz2==) return;
         if(max(tree[tree[x].son[]].sz1,tree[tree[x].son[]].sz2)
             >max(tree[tree[x].son[]].sz1,tree[tree[x].son[]].sz2)){
                 tree[x].son[]=merge(tree[x].son[],tree[x].son[]);
                 tree[x].son[]=;
                 solve(tree[x].son[],key+(<<p-),p-);
             }else{
                 tree[x].son[]=merge(tree[x].son[],tree[x].son[]);
                 tree[x].son[]=;
                 solve(tree[x].son[],key+(<<p-),p-);
             }
         up(x);
     }
 }seg;
 int main(){
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     int T=read();
         while(T--){
         seg.clear(); ans.clear();
         int n=read();
         for(int i=;i<=n;++i)
         {
             int x=read();
             seg.insert(x,,);
         }

         for(int i=;i<=n;++i)
         {
             int x=read();
             seg.insert(x,,);
         }

         seg.solve(,,);
         sort(ans.begin(),ans.end());
         for(int i=;i<ans.size()-;++i) printf("%d ",ans[i]);
         printf("%d\n",ans[ans.size()-]);
     }
     return ;
 }

做法二：

 #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  100010
 #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 int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int inf=0x7fffffff;
       int dx[]={-,,,};
       int dy[]={,,-,};

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

 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)
         {
             ans[++m]=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],,);
     }

     m=;
     while(m<n)
     {
         int x=find();
         dfs(x,,-);
     }

     sort(ans+,ans+n+);
     rep(i,,n-) printf("%d ",ans[i]);
     printf("%d\n",ans[n]);
     rep(i,,cnt)
      rep(j,,) t[i][j]=s[i][j]=;
 }

 int main()
 {
     int cas=read();
     while(cas--) solve();
     return ;
 }