异或和之差

异或和之差-原题链接
对于大佬的题解进行一部分说明

import java.io.*;
import java.util.*;

public class Main {
    
    static int N = 200010, M = 21 * N;
    static int son[][] = new int[M][2], idx;
    // l[i]表示从 i到n 范围内异或和子段的最大/最小值
    static int lmin[] = new int[N], lmax[] = new int[N];
    // r[i]表示从 1到i 范围内异或和子段的最大/最小值
    static int rmin[] = new int[N], rmax[] = new int[N];
    static int a[] = new int[N];
    static int n;
    
    /* 
        a^b^b = a
        所以对于区间也是如此: 
        比如: 在计算rmin[]和rmax[]时, 当枚举到k时
        sum是一段连续的区间异或和: a[1]^a[2]...^a[k],
        每次query_min(sum)时, 因为trie里面存的是每个从1到k的区间异或和:[1-3], [1-6]...
        比如 sum ^ [1-3] = [4-k],
        所以每次异或后求得的最小值即为区间1到k中的某个字段的异或和, 
        求一下前缀, 求一下后缀, 最后计算一下, 就能覆盖所有的情况
    */
    public static void insert(int x){
        int p = 0;
        for(int i = 20; i >= 0; i--){
            int u = x >> i & 1;
            if(son[p][u] == 0) son[p][u] = ++idx;
            p = son[p][u];
        }
    }
    
    public static int query_min(int x){
        int p = 0, res = 0;
        for(int i = 20; i >= 0; i--){
            int u = x >> i & 1;
            if(son[p][u] != 0){
                p = son[p][u];
            }else{
                p = son[p][u ^ 1];
                res |= 1 << i;
            }
        }
        return res;
    }
    
    public static int query_max(int x){
        int p = 0, res = 0;
        for(int i = 20; i >= 0; i--){
            int u = x >> i & 1;
            if(son[p][u ^ 1] != 0){
                p = son[p][u ^ 1];
                res |= 1 << i;
            }else {
                p = son[p][u];
            }
            
        }
        return res;
    }
    
    public static void solve(){
        // 计算前缀1-r
        int sum = 0;
        rmin[0] = (int)2e9;
        rmax[0] = -1;
        insert(0);
        for(int i = 1; i <= n; i++){
            sum ^= a[i];
            rmin[i] = Math.min(rmin[i - 1], query_min(sum));
            rmax[i] = Math.max(rmax[i - 1], query_max(sum));
            insert(sum);
        }
        
        for(int i = 0; i < M; i++) Arrays.fill(son[i], 0);
        idx = 0;
        // 计算后缀l-n
        sum = 0;
        lmin[n + 1] = (int)2e9;
        lmax[n + 1] = -1;
        insert(0);
        for(int i = n; i >= 1; i--){
            sum ^= a[i];
            lmin[i] = Math.min(lmin[i + 1], query_min(sum));
            lmax[i] = Math.max(lmax[i + 1], query_max(sum));
            insert(sum);
        }
        
        int ans = 0;
        for(int i = 1; i < n; i++){
            ans = Math.max(ans, Math.max(lmax[i + 1] - rmin[i], rmax[i] - lmin[i + 1]));
        }
        System.out.println(ans);
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StreamTokenizer stmInput = new StreamTokenizer(br);
        stmInput.nextToken();
        n = (int)stmInput.nval;
        for(int i = 1; i <= n; i++){
            stmInput.nextToken();
            a[i] = (int)stmInput.nval;
        }
        solve();
        br.close();
    }
}
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