【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

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef pair<int,int> PII;
  7 typedef pair<ll,ll> Pll;
  8 typedef vector<int> VI;
  9 typedef vector<PII> VII;
 10 //typedef pair<ll,ll>P;
 11 #define N  200010
 12 #define M  200010
 13 #define fi first
 14 #define se second
 15 #define MP make_pair
 16 #define pb push_back
 17 #define pi acos(-1)
 18 #define mem(a,b) memset(a,b,sizeof(a))
 19 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 20 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 21 #define lowbit(x) x&(-x)
 22 #define Rand (rand()*(1<<16)+rand())
 23 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 24 #define ls p<<1
 25 #define rs p<<1|1
 26 
 27 const ll MOD=1e9+7,inv2=(MOD+1)/2;
 28       double eps=1e-6;
 29       ll INF=1e15;
 30       int dx[4]={-1,1,0,0};
 31       int dy[4]={0,0,-1,1};
 32 
 33 int t[N*32][2],s[N*32][2],a[N],b[N],m,cnt;
 34 ll ans;
 35 
 36 int read()
 37 {
 38    int v=0,f=1;
 39    char c=getchar();
 40    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 41    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 42    return v*f;
 43 }
 44 
 45 void update(int x,int y,int op)
 46 {
 47     int u=1;
 48     per(i,30,0)
 49     {
 50         int now=(x>>i)&1;
 51         if(!t[u][now]) t[u][now]=++cnt;
 52         u=t[u][now];
 53         s[u][op]+=y;
 54     }
 55 }
 56 
 57 int query(int x,int op)
 58 {
 59     int u=1,res=0;
 60     per(i,30,0)
 61     {
 62         int now=(x>>i)&1;
 63         if(t[u][now^1]&&s[t[u][now^1]][op])
 64         {
 65             if(now^1) res+=1<<i;
 66             u=t[u][now^1];
 67         }
 68          else
 69          {
 70              if(now) res+=1<<i;
 71              u=t[u][now];
 72          }
 73 
 74     }
 75     return res;
 76 }
 77 
 78 int find(int op)
 79 {
 80     int u=1,res=0;
 81     per(i,30,0)
 82     {
 83         if(t[u][0]&&s[t[u][0]][op]) u=t[u][0];
 84          else
 85          {
 86              res+=1<<i;
 87              u=t[u][1];
 88          }
 89     }
 90     return res;
 91 }
 92 
 93 int dfs(int x,int op,int pre)
 94 {
 95     while(1)
 96     {
 97         int y=query(x,op^1);
 98         if(y==pre)
 99         {
100             m++;
101             ans+=(x^y);
102             update(x,-1,op);
103             update(y,-1,op^1);
104             return 1;
105         }
106         if(dfs(y,op^1,x)) return 0;
107     }
108 }
109 
110 void solve()
111 {
112     int n=read();
113     cnt=1;
114     rep(i,1,n)
115     {
116         a[i]=read();
117         update(a[i],1,0);
118     }
119     rep(i,1,n)
120     {
121         b[i]=read();
122         update(b[i],1,1);
123     }
124     ans=0;
125     m=0;
126     while(m<n)
127     {
128         int x=find(0);
129         dfs(x,0,-1);
130     }
131     printf("%I64d\n",ans);
132     rep(i,1,cnt)
133      rep(j,0,1) t[i][j]=s[i][j]=0;
134 }
135 
136 int main()
137 {
138     //freopen("1.in","r",stdin);
139     int cas=read();
140     while(cas--) solve();
141     return 0;
142 }