CF-282-E-Trie

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-232.CF-87-D-并查集2024-01-133.CF-91-B-单调栈+二分2024-01-234.CF-240-F-线段树2024-01-215.CF-242-E-线段树2024-01-206.CF-292-D-并查集2024-01-23

7.CF-282-E-Trie2024-01-17

8.CF-342-E-分块2024-01-129.CF-375-D-启发式合并2024-01-1210.CF-431-D-二分+数位DP2024-01-2411.CF-461-B-树形DP2024-01-2212.CF-514-D-单调队列2024-01-1413.CF-522-D-线段树2024-05-0514.CF-570-D-启发式合并2024-01-2115.CF-600-E-启发式合并2024-05-0416.CF-613-D-虚树2024-01-1317.CF-620-E-DFS序+线段树2024-01-1718.CF-675-E-RMQ优化DP2024-01-1619.CF-702-E-倍增2024-01-1720.CF-797-E-根号分治2024-05-0421.CF-817-E-Trie2024-01-1622.CF-877-E-dfs序+线段树2024-05-0523.CF-915-F-并查集2024-01-2124.CF-938-D-最短路2024-05-1025.CF-1051-F-最短路+最小生成树2024-01-1726.CF-1184-E3-最小生成树+倍增+并查集2024-01-2427.CF-1304-E-倍增LCA+思维2024-01-2528.CF-1399-E2-优先队列2024-01-2229.CF-1515-F-思维2024-02-2430.CF-1831-E-卡特兰数+异或哈希+差分2024-01-2231.CF-1921-F-根号分治2024-01-24

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282-E 题目大意

给定一个长为$n$的序列$a$，要求选一个前缀与一个不相交的后缀，求这些元素异或结果的最大值。

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Solution

最大的异或值，首先就要想到$Trie$。

把所有前缀加入$Trie$，然后枚举后缀，一边在$Trie$中查询最大异或值，一边删掉一个前缀，避免会产生相交的前后缀。

注意要spj掉前缀为空与后缀为空的特殊情况，时间复杂度$O(nlogn)$。

#include<bits/stdc++.h>
using namespace std;
using ll=long long;

const int N=1e5+10;
int ch[N*61][2],cnt[N*31];
int idx=0;

void update(ll x,int k){
    int p=0;
    for(int i=60;~i;i--){
        int j=(x>>i)&1;
        if(!ch[p][j]) ch[p][j]=++idx;
        p=ch[p][j];
        cnt[p]+=k;
    }
}

ll query(ll x){
    int p=0;
    ll res=0;
    for(int i=60;~i;i--){
        int j=(x>>i)&1;
        if(cnt[ch[p][!j]]){
            res|=(1LL<<i);
            p=ch[p][!j];
        }else{
            p=ch[p][j];
        }
    }
    return res;
}

void solve(){
    int n;
    cin>>n;
    ll pre=0,suf=0;
    vector<ll> a(n+1);
    update(0,1);
    for(int i=1;i<=n;i++){
        cin>>a[i];
        pre^=a[i];
        update(pre,1);
    }
    ll ans=query(0);
    for(int i=n;~i;i--){
        suf^=a[i];
        update(pre,-1);
        pre^=a[i];
        ans=max(ans,query(suf));
    }
    cout<<ans<<'\n';
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}