Bzoj3689 exclusive or

N non-negative integers A [1], a [2],..., A [n].
For each pair (I, j) Satisfying 1 <= I <j <= n, a new number A [I] XOR A [J] is obtained. in this case, there are N * (n-1)/two new numbers. Evaluate the first K of these numbers (excluding a [I.

Note: XOR corresponds to "XOR" in Pascal and "^" in C ++ ".

Idea: Same as noi2010 super piano. Http://blog.csdn.net/wyfcyx_forever/article/details/40400327

We only need to first put each number in the global heap and the minimum value after the number exclusive, and then put the k + 1 small value in the heap every time the heap goes out. This repeat K times.

Then the problem becomes to quickly find the K small value of XOR for a given number in a certain range. We can maintain the size by using the persistent trie and directly divide it into two points on the tree.

Time complexity is still O (klogn ).

Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std; #define N 100010int w[N]; int ch[N<<5][2], size[N<<5], ind;int root[N];int Newadd(int Last, int bit, int x) {    int q = ++ind;    ch[q][0] = ch[Last][0], ch[q][1] = ch[Last][1];    size[q] = size[Last] + 1;         if (bit < 0)        return q;         if ((x >> bit) & 1)        ch[q][1] = Newadd(ch[Last][1], bit - 1, x);    else        ch[q][0] = Newadd(ch[Last][0], bit - 1, x);    return q;}int getkth(int Left, int Right, int x, int k) {    int res = 0, better;    bool d;    for(int dep = 30; dep >= 0; --dep) {        d = (x >> dep) & 1;        better = size[ch[Right][d]] - size[ch[Left][d]];        if (better >= k)            Left = ch[Left][d], Right = ch[Right][d];        else            Left = ch[Left][d ^ 1], Right = ch[Right][d ^ 1], res += (1<<dep), k -= better;    }         return res;} struct st {    int end, kth, val;    st(int _end = 0, int _kth = 0, int _val = 0):end(_end),kth(_kth),val(_val){}    bool operator < (const st &B) const {        return val < B.val;    }}; #define K 250010struct Heap {    st a[K];    int top;    Heap():top(0){}    inline void up(int x) {        for(; x != 1; x >>= 1)            if (a[x]<a[x>>1])                swap(a[x],a[x>>1]);            else                break;    }    inline void down(int x) {        int son;        for(; (x<<1)<=top; ) {            son=(((x<<1)==top)||(a[x<<1]<a[(x<<1)|1]))?(x<<1):((x<<1)|1);            if (a[son]<a[x])                swap(a[son],a[x]),x=son;            else                break;        }    }    inline void push(const st &x) {        a[++top] = x;        up(top);    }    st Min() {        return a[1];    }    inline void pop() {        a[1] = a[top--];        down(1);    }}H; int main() {    int n, k;    scanf("%d%d", &n, &k);         register int i, j;    for(i = 1; i <= n; ++i) {        scanf("%d", &w[i]);        root[i] = Newadd(root[i - 1], 30, w[i]);    }         for(i = 2; i <= n; ++i)        H.push(st(i, 1, getkth(root[0], root[i - 1], w[i], 1)));         for(i = 1; i <= k; ++i) {        st tmp = H.Min();        H.pop();        if (i != 1)            putchar(' ');        printf("%d", tmp.val);        if (tmp.kth != tmp.end - 1)            H.push(st(tmp.end, tmp.kth + 1, getkth(root[0], root[tmp.end - 1], w[tmp.end], tmp.kth + 1)));    }         return 0;}

Bzoj3689 exclusive or