P5283 [十二省联考2019]异或粽子

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求最大第$k$个异或区间和。

维护每个右端点，$r-[0,l-1]$，放入堆中维护，取出来后分裂成两个$r-[0,p),(p,l-1]$继续放入堆中维护即可

  • $2^{32}$>int
  • 可以选择二分查找最大值所位置,也可以可持续字典树的结点上返回位置。
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef unsigned int ll;

#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;

ll read()
{
    ll x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
int rt[N];
struct Tri
{
    int scnt, ch[N * 33][2], cnt[N * 33];
    int q[N * 32];
    void insert(int &pos, int pre, ll x, int id)
    {
        pos = ++scnt;
        int tmp = pos;
        cnt[tmp] = cnt[pre] + 1;
        for (int i = 32; i >= 0; i--)
        {
            int c = (x & (1 << i)) >> i;

            ch[tmp][1] = ch[pre][1];
            ch[tmp][0] = ch[pre][0];
            ch[tmp][c] = ++scnt;

            tmp = ch[tmp][c], pre = ch[pre][c];
            cnt[tmp] = cnt[pre] + 1;
        }
        q[tmp] = id;
    }
    int query(ll x, int l, int r)
    {

        l = (l == 0) ? 0 : rt[l - 1];
        r = rt[r];
        for (int i = 32; i >= 0; i--)
        {
            int c = (x & (1 << i)) >> i;
            if ((cnt[ch[r][c ^ 1]] - cnt[ch[l][c ^ 1]]) > 0)
            {
                r = ch[r][c ^ 1];
                l = ch[l][c ^ 1];
            }
            else
            {
                r = ch[r][c];
                l = ch[l][c];
            }
        }
        return q[r];
    }
} T;

ll a[N], s[N];

struct node
{
    int l, r;
    ll v, maxx;
    int mid;
    friend bool operator<(node A, node B)
    {
        return A.maxx < B.maxx;
    }
};

int main()
{
    int n = read(), k = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    for (int i = 1; i <= n; i++)
        s[i] = s[i - 1] ^ a[i];
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        T.insert(rt[i], i == 0 ? 0 : rt[i - 1], s[i], i);
    }
    priority_queue<node> q;
    for (int i = 1; i <= n; i++)
    {
        int id = T.query(s[i], 0, i - 1);
        q.push((node){0, i - 1, s[i], s[i] ^ s[id], id});
    }
    ll ans = 0;
    for (int i = 1; i <= k; i++)
    {
        node now = q.top();
        q.pop();
        ans += now.maxx;
        int mid = now.mid;
        if (mid - 1 >= now.l)
        {
            int id = T.query(now.v, now.l, mid - 1);
            q.push((node){now.l, mid - 1, now.v, now.v ^ s[id], id});
        }
        if (now.r >= mid + 1)
        {
            int id = T.query(now.v, mid + 1, now.r);
            q.push((node){mid + 1, now.r, now.v, now.v ^ s[id], id});
        }
    }
    printf("%lld\n", ans);
}