BZOJ 3689: 异或之

字典树可以$o(logn)查找第k大$

使用$可持久化Trie 区间查找第k大，然后首先把每个数异或之后的最小丢进小根堆中，然后一个一个取出，取出后就再丢次小，一共取k次$

总的时间复杂度为$O(klogn)$

本来的考虑是 先找出第k大，然后在$Trie上DFS把小于这个数的全丢进vector  然后发现会有很多无用状态会搜索到，T掉$

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 int n, k, arr[N], kth;
 struct node
 {
     int base, num, pos, l, r;
     node () {}
     node (int base, int num, int pos, int l, int r) : base(base), num(num), pos(pos), l(l), r(r) {}
     bool operator < (const node &r) const { return num > r.num; }
 };
 priority_queue <node> q;  

 namespace TRIE
 {
     int rt[N], cnt;
     struct node
     {
         int son[], cnt;
     }a[N * ];
     void init()
     {
         a[].son[] = a[].son[] = a[].cnt = ;
         cnt = ;
     }
     void insert(int &now, int pre, int x)
     {
         now = ++cnt;
         a[now] = a[pre];
         int root = now; ++a[root].cnt;
         for (int i = ; i >= ; --i)
         {
             int id = (x >> i) & ;
             a[++cnt] = a[a[root].son[id]];
             ++a[cnt].cnt;
             a[root].son[id] = cnt;
             root = cnt;
         }
     }
     int find_kth(int x, int k, int l, int r)
     {
         int res = ;
         int tl = rt[l], tr = rt[r];
         for (int s = ; s >= ; --s)
         {
             int id = (x >> s) & ;
             int sum = a[a[tr].son[id]].cnt - a[a[tl].son[id]].cnt;
             if (sum < k)
             {
                 k -= sum;
                 res |=  << s;
                 tr = a[tr].son[id ^ ];
                 tl = a[tl].son[id ^ ];
             }
             else
             {
                 tr = a[tr].son[id];
                 tl = a[tl].son[id];
             }
         }
         return res;
     }
 }

 int main()
 {
     while (scanf("%d%d", &n, &k) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d", arr + i);
         TRIE::init();
         for (int i = ; i <= n; ++i) TRIE::insert(TRIE::rt[i], TRIE::rt[i - ], arr[i]);
         for (int i = ; i < n; ++i) q.push(node(arr[i], TRIE::find_kth(arr[i], , i, n), , i, n));
         for (int i = ; i <= k; ++i)
         {
             node top = q.top(); q.pop();
             printf("%d%c", top.num, " \n"[i == k]);
             q.push(node(top.base, TRIE::find_kth(top.base, top.pos, top.l, top.r), top.pos + , top.l, top.r));
         }
     }
     return ;
 }