【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=5+5e6;

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;
}

vector<int> ans;
struct SEG{
    struct Node{
        int son[2],sz1,sz2;
    }tree[MAX_N];
    inline int new_node(){
        tree[++top]=(Node){{0,0},0,0};
        return top;
    }
    int rt,top;
    inline void up(int x){
        tree[x].sz1=tree[tree[x].son[0]].sz1+tree[tree[x].son[1]].sz1;
        tree[x].sz2=tree[tree[x].son[0]].sz2+tree[tree[x].son[1]].sz2;
    }
    void clear(){ top=0,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=30;i>=0;--i){
            bool t=pos&(1<<i);
            if(tree[x].son[t]==0)
                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==0||y==0) return x+y;
        tree[x].sz1+=tree[y].sz1;
        tree[x].sz2+=tree[y].sz2;
        tree[x].son[0]=merge(tree[x].son[0],tree[y].son[0]);
        tree[x].son[1]=merge(tree[x].son[1],tree[y].son[1]);
        return x;
    }
    void solve(int x,int key,int p){
//        printf("{%d %d %d}",x,key,p);
        if(p==0){
            int t=min(tree[x].sz1,tree[x].sz2);
//            printf("[%d]",t);
            for(int i=1;i<=t;++i) ans.push_back(key);
            tree[x].sz1-=t;
            tree[x].sz2-=t;
            return;
        }
        if(tree[x].sz1==0||tree[x].sz2==0) return;
        solve(tree[x].son[0],key,p-1);
        solve(tree[x].son[1],key,p-1);
        up(x);
        if(tree[x].sz1==0||tree[x].sz2==0) return;
        if(max(tree[tree[x].son[0]].sz1,tree[tree[x].son[0]].sz2)
            >max(tree[tree[x].son[1]].sz1,tree[tree[x].son[1]].sz2)){
                tree[x].son[0]=merge(tree[x].son[0],tree[x].son[1]);
                tree[x].son[1]=0;
                solve(tree[x].son[0],key+(1<<p-1),p-1);
            }else{
                tree[x].son[1]=merge(tree[x].son[0],tree[x].son[1]);
                tree[x].son[0]=0;
                solve(tree[x].son[1],key+(1<<p-1),p-1);
            }
        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=1;i<=n;++i)
        {
            int x=read();
            seg.insert(x,1,0);
        }

        for(int i=1;i<=n;++i)
        {
            int x=read();
            seg.insert(x,0,1);
        }

        seg.solve(1,0,31);
        sort(ans.begin(),ans.end());
        for(int i=0;i<ans.size()-1;++i) printf("%d ",ans[i]);
        printf("%d\n",ans[ans.size()-1]);
    }
    return 0;
}

 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+7,inv2=(MOD+1)/2;
      double eps=1e-4;
      int INF=1e9;
      int inf=0x7fffffff;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

int t[N*31][2],s[N*31][2],a[N],b[N],ans[N],m,cnt;

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,29,0)
    {
        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,29,0)
    {
        int now=(x>>i)&1;
        if(t[u][now]&&s[t[u][now]][op])
        {
            if(now) res+=1<<i;
            u=t[u][now];
        }
         else
         {
             if(now^1) res+=1<<i;
             u=t[u][now^1];
         }

    }
    return res;
}

int find(int op)
{
    int u=1,res=0;
    per(i,29,0)
    {
        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)
        {
            ans[++m]=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);
    }

    m=0;
    while(m<n)
    {
        int x=find(0);
        dfs(x,0,-1);
    }

    sort(ans+1,ans+n+1);
    rep(i,1,n-1) printf("%d ",ans[i]);
    printf("%d\n",ans[n]);
    rep(i,1,cnt)
     rep(j,0,1) t[i][j]=s[i][j]=0;
}

int main()
{
    int cas=read();
    while(cas--) solve();
    return 0;
}