[十二省聯考2019]異或糉子（堆+可持久化Trie）

前置芝士：可持久化Trie & 堆

相似於超級鋼琴，咱們用堆維護一個四元組\((st, l, r, pos)\)表示以\(st\)爲起點，終點在\([l, r]\)內，裏面的最大值的位置爲\(pos\)

咱們維護一個小根堆（堆頂最大），權值爲st-pos的異或和，每一次找出最大的並刪掉

所謂刪，就是把一個區間從pos處分裂

即：\((st, l, r)->(st, l, pos - 1) (st, pos + 1, r)\)

這樣從新維護pos值便可

維護pos值時，咱們須要維護區間內與x的異或值最大，不難想到可持久化\(Trie\)，因而只須要把超級鋼琴中的\(RMQ\)變成可持久化\(Trie\)便可

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define re register
#define debug printf("Now is Line : %d\n",__LINE__)
#define file(a) freopen(#a".in","r",stdin);freopen(#a".out","w",stdout)
#define ll long long
il ll read() {
    re ll x = 0, f = 1; re char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
    return x * f;
}
#define rep(i, s, t) for(re int i = s; i <= t; ++ i)
#define maxn 500005
int n, m;
ll sum[maxn], ans;
namespace Trie {
    struct PAX {
        int id, s, ch[2];
    }e[maxn * 50];
    int cnt, rt[maxn];
    il void insert(int&k, int kk, int bit, int id, ll val) {
        k = ++ cnt;
        e[k] = e[kk], ++ e[k].s;
        if(bit == -1) return(void)(e[k].id = id);
        int c = (val >> bit) & 1;
        insert(e[k].ch[c], e[kk].ch[c], bit - 1, id, val); 
    }
    il int query(int l, int r, int bit, ll val) {
        if(bit == -1) return e[r].id;
        int c = (val >> bit) & 1;
        if(e[e[r].ch[c ^ 1]].s > e[e[l].ch[c ^ 1]].s) 
            return query(e[l].ch[c ^ 1], e[r].ch[c ^ 1], bit - 1, val);
        return query(e[l].ch[c], e[r].ch[c], bit - 1, val);
    }
}
using namespace Trie;
struct node {
    int st, l, r, pos;
    il bool operator < (const node a) const {
        return (sum[pos] ^ sum[st - 1]) < (sum[a.pos] ^ sum[a.st - 1]);
    }
    node(int St, int L, int R) {
        st = St, l = L, r = R, pos = query(rt[l - 1], rt[r], 32, sum[st - 1]);
    }
};
priority_queue<node>q;
signed main() {
    file(a);
    n = read(), m = read();
    rep(i, 1, n) sum[i] = sum[i - 1] ^ read();
    rep(i, 1, n) insert(rt[i], rt[i - 1], 32, i, sum[i]);
    rep(i, 1, n) q.push(node(i, i, n));
    while(m --) {
        node t = q.top(); q.pop();
        ans += sum[t.pos] ^ sum[t.st - 1];
        if(t.l < t.pos) q.push(node(t.st, t.l, t.pos - 1));
        if(t.pos < t.r) q.push(node(t.st, t.pos + 1, t.r));
    }
    printf("%lld", ans);
    return 0;
}