P4585 [FJOI2015]火星商店问题
P4585 [FJOI2015]火星商店问题
线段树分治,按时间分治,将在时间区间内的客人加入线段树。
要求异或和最大,考虑可持久化 trie
维护。
将商品按编号排序,便于询问时二分查找,再根据时间分治即可,满足任何时刻商品编号递增。
时间复杂度 。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ha putchar(' ')
#define he putchar('\n')
inline int read()
{
int 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 << 3) + (x << 1) + (c ^ 48), c = getchar();
return x * f;
}
inline void write(int x)
{
if(x < 0)
{
putchar('-');
x = -x;
}
if(x > 9)
write(x / 10);
putchar(x % 10 + 48);
}
const int _ = 1e5 + 10;
int n, m, tot, rt[_], t1, t2, sz[_ * 20], c[_ * 20][2], ans[_];
vector<int> a[_ << 2];
int top, st[_];
struct guest
{
int l, r, L, R, x;
} p[_];
struct buy
{
int s, v, t;
} q[_], k1[_], k2[_];
bool cmp(const buy x, const buy y) {return x.s < y.s;}
void ins(int &x, int u, int w)
{
int nw; nw = x = ++tot;
for(int i = 17; i >= 0; --i)
{
bool d = w & (1 << i);
c[nw][d ^ 1] = c[u][d ^ 1], c[nw][d] = ++tot;
nw = c[nw][d], u = c[u][d];
sz[nw] = sz[u] + 1;
}
}
int qry(int l, int r, int w)
{
int res = 0;
for(int i = 17; i >= 0; --i)
{
bool d = w & (1 << i);
if(sz[c[r][d ^ 1]] - sz[c[l][d ^ 1]] > 0) l = c[l][d ^ 1], r = c[r][d ^ 1], res += (1 << i);
else l = c[l][d], r = c[r][d];
}
return res;
}
void upd(int o, int l, int r, int L, int R, int x)
{
if(L > R || r < L || l > R) return;
if(L <= l && r <= R) return a[o].push_back(x), void();
int mid = (l + r) >> 1;
if(L <= mid) upd(o << 1, l, mid, L, R, x);
if(R > mid) upd(o << 1 | 1, mid + 1, r, L, R, x);
}
void solve(int o, int l, int r, int L, int R)
{
if(L > R) return;
top = tot = 0;
for(int i = L; i <= R; ++i)
{
st[++top] = q[i].s;
ins(rt[top], rt[top - 1], q[i].v);
}
for(int i = 0; i < a[o].size(); ++i)
{
int k = a[o][i];
int l = upper_bound(st + 1, st + top + 1, p[k].l - 1) - st - 1;
int r = upper_bound(st + 1, st + top + 1, p[k].r) - st - 1;
ans[k] = max(ans[k], qry(rt[l], rt[r], p[k].x));
}
int c1 = 0, c2 = 0;
if(l == r) return;
int mid = (l + r) >> 1;
for(int i = L; i <= R; ++i)
if(q[i].t <= mid) k1[++c1] = q[i];
else k2[++c2] = q[i];
for(int i = 1; i <= c1; ++i) q[i + L - 1] = k1[i];
for(int i = 1; i <= c2; ++i) q[i + L - 1 + c1] = k2[i];
solve(o << 1, l, mid, L, L + c1 - 1), solve(o << 1 | 1, mid + 1, r, L + c1, R);
}
signed main()
{
n = read(), m = read();
for(int i = 1; i <= n; ++i) ins(rt[i], rt[i - 1], read());
for(int i = 1, opt, l, r, x, d, s, v; i <= m; ++i)
{
opt = read();
if(!opt)
{
s = read(), v = read();
q[++t1] = {s, v, t1};
}
else
{
l = read(), r = read(), x = read(), d = read();
ans[++t2] = qry(rt[l - 1], rt[r], x);
p[t2] = {l, r, max(1, t1 - d + 1), t1, x};
}
}
for(int i = 1; i <= t2; ++i) upd(1, 1, t1, p[i].L, p[i].R, i);
sort(q + 1, q + t1 + 1, cmp);
solve(1, 1, t1, 1, t1);
for(int i = 1; i <= t2; ++i) printf("%d\n", ans[i]);
return 0;
}
本文来自博客园,作者:蒟蒻orz,转载请注明原文链接:https://www.cnblogs.com/orzz/p/18121977