BZOJ4260: Codechef REBXOR (01Tire树)

题意

题目链接

sol

首先维护出前缀xor和后缀xor

对每个位置的元素插入到trie树里面，每次找到和该前缀xor起来最大的元素

正反各做一遍，取最大。

记得要开log倍空间qwq。。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 4e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-')f =- 1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, a[maxn], s[maxn], pmx[maxn], smx[maxn], ch[maxn][2], tot;
void insert(int x) {
    int now = 0;
    for(int i = 0; i <= 30; i++) {
        int nxt = (x >> i & 1);
        if(!ch[now][nxt]) ch[now][nxt] = ++tot;
        now = ch[now][nxt];
    }
}
int query(int x) {
    int now = 0, ans = 0;
    for(int i = 0; i <= 30; i++) {
        int nxt = (x >> i & 1);
        if(ch[now][nxt ^ 1]) ans += 1 << i, now = ch[now][nxt ^ 1];
        else now = ch[now][nxt];
    }
    return ans;
}
void solve(int *s, int *mx) {
    for(int i = 1; i <= n; i++) {
        s[i] = s[i - 1] ^ a[i];
        insert(s[i]);
        mx[i] = query(s[i]);
        mx[i] = max(mx[i - 1], mx[i]);
    }
}
int main() {
    //freopen("3.in", "r", stdin); 
    n = read();
    for(int i = 1; i <= n; i++) a[i] = read();
    solve(s, pmx);  
    reverse(a + 1, a + n + 1);
    solve(s, smx);
    reverse(smx + 1, smx + n + 1);
    int ans = 0;
    for(int i = 1; i <= n; i++) ans = max(ans, pmx[i] + smx[i + 1]);
    cout << ans;
}