[BZOJ 3261]最大异或和
Description
给定一个初始长度为 $N$ 的非负整数序列 ${A}$ 。有 $M$ 个操作,支持:
- $A~x$ :添加操作,表示在序列末尾添加一个数 $x$ ,序列的长度 $N+1$ 。
- $Q~l~r~x$ :询问操作,求 $$\max_{l\leq p\leq r}A_p\oplus A_{p+1}\oplus\cdots\oplus A_N\oplus x$$
$1\leq N,M\leq 300000,0\leq A_i\leq 10^7$
Solution
不妨记 $A_i$ 的前缀异或和为 $B_i$ ,显然题目转化为 $$\max_{l\leq p\leq r}B_{p-1}\oplus B_N\oplus x$$
我们不妨用类似主席树的思路,我们建一棵 $B_i$ 的 $Trie$ ,并且将它持久化,然后贪心的选择。显然是可行的。
值得注意的是由于异或的前缀和,并且又因为差分,查询时的区间应该为 $[l-2,r-1]$ ,为了防止越界,可以先插入个 $0$ 进 $Trie$ 。
Code
//It is made by Awson on 2018.3.16
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 300000;
void read(int &x) {
char ch; bool flag = 0;
for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }
int n, m, bin[30], a, b, t, A[30], l, r; char ch[5];
struct Trie {
int root[(N<<1)+5], ch[N*50+5][2], key[N*50+5], pos;
int cpynode(int o) {++pos; ch[pos][0] = ch[o][0], ch[pos][1] = ch[o][1], key[pos] = key[o]; return pos; }
void insert(int &o, int *A) {
int u = o = cpynode(o); ++key[u];
for (int i = 25; i >= 1; i--) {
ch[u][A[i]] = cpynode(ch[u][A[i]]);
u = ch[u][A[i]], ++key[u];
}
}
int query(int a, int b, int *A) {
int ans = 0;
for (int i = 25; i >= 1; i--) {
if (key[ch[b][A[i]^1]]-key[ch[a][A[i]^1]] > 0) ans += bin[i-1], a = ch[a][A[i]^1], b = ch[b][A[i]^1];
else a = ch[a][A[i]], b = ch[b][A[i]];
}
return ans;
}
}T;
void work() {
read(n), read(m); bin[0] = 1; for (int i = 1; i <= 25; i++) bin[i] = (bin[i-1]<<1);
T.insert(T.root[1], A); ++n;
for (int i = 2; i <= n; i++) {
read(a), b ^= a, t = b; for (int j = 1; j <= 25; j++) A[j] = t%2, t /= 2;
T.insert(T.root[i] = T.root[i-1], A);
}
while (m--) {
scanf("%s", ch);
if (ch[0] == 'A') {
++n; read(a), b ^= a, t = b;
for (int j = 1; j <= 25; j++) A[j] = t%2, t /= 2;
T.insert(T.root[n] = T.root[n-1], A);
}else {
read(l), read(r), read(t); t ^= b;
for (int j = 1; j <= 25; j++) A[j] = t%2, t /= 2;
writeln(T.query(T.root[l-1], T.root[r], A));
}
}
}
int main() {
work(); return 0;
}
博主蒟蒻,随意转载。但必须附上原文链接:http://www.cnblogs.com/NaVi-Awson/,否则你会终生找不到妹子!!!