01 Trie 专题

异或最大值

The xor largest pair

题意：异或最大值的模板。一个数和一个序列中一个数的异或最大值是多少？要支持询问。

思路:考虑把序列插入，构建一个 $\text{Trie}$ 树。那么在询问时，只需要讨论该数的位是 $0$ 还是 $1$ 就行了。这样就需要 $O(n\log w)$ 的预处理，$O(\log w)$ 的询问和修改，为什么是对的。因为异或中我们如果可以满足最高位，那么没有理由不改变最高位，因为 $2^{bit} >\sum_{i=0}^{i < bit}2^i$ 。那么由高位到地位贪心就可以了。

/*
https://ac.nowcoder.com/acm/problem/50993
*/
#include<bits/stdc++.h>
using namespace std;

const int maxn = 2e6 + 100;
int ch[maxn][2];

int sz = 0, rt, n;

int read() {int x; scanf("%d", &x); return x;}

void insert(int u, int t, int x){
    if(t < 0) return;
    int f = ((x >> t) & 1);
    if(!ch[u][f]) ch[u][f] = ++sz;
    insert(ch[u][f], t-1, x);
}

int ask(int u, int t, int x){
    if(t < 0) return 0;
    int f = ((x >> t) & 1);
    if(ch[u][!f]) return ask(ch[u][!f], t-1, x) + (1 << t);
    
    return ask(ch[u][f], t-1, x);
}

int main(){
    n = read();
    int ans = 0;
    rt = ++ sz;
    for(int i = 0; i < n; i++){
        int x = read();
        insert(rt, 30, x);
        if(i != 0) ans = max(ans, ask(rt, 30, x));
    }
    printf("%d\n", ans);
    return 0;
}

Xor Sum

Problem Description

Zeus 和 Prometheus 做了一个游戏，Prometheus 给 Zeus 一个集合，集合中包含了N个正整数，随后 Prometheus 将向 Zeus 发起M次询问，每次询问中包含一个正整数 S ，之后 Zeus 需要在集合当中找出一个正整数 K ，使得 K 与 S 的异或结果最大。Prometheus 为了让 Zeus 看到人类的伟大，随即同意 Zeus 可以向人类求助。你能证明人类的智慧么？

Input

输入包含若干组测试数据，每组测试数据包含若干行。
输入的第一行是一个整数$T（T < 10）$，表示共有T组数据。
每组数据的第一行输入两个正整数$N，M（<1=N,M<=100000）$，接下来一行，包含N个正整数，代表 Zeus 的获得的集合，之后M行，每行一个正整数S，代表 Prometheus 询问的正整数。所有正整数均不超过$2^{32}$。

Output

对于每组数据，首先需要输出单独一行”Case #?:”，其中问号处应填入当前的数据组数，组数从1开始计算。
对于每个询问，输出一个正整数K，使得K与S异或值最大。

Sample Input

Sample Output

Case #1:
4
3
Case #2:
4

同上题，简单板子

代码

/*
http://acm.hdu.edu.cn/showproblem.php?pid=4825
*/
#include<bits/stdc++.h>
using namespace std;

int read(){int x; scanf("%d", &x); return x;}

const int maxn = 5e6 + 10;
int ch[maxn][2], value[maxn];
int a[maxn];
int sz = 1, rt = 1;
int m, n, t, T;

void insert(int u, int t, int x){
    if(t < 0) { value[u] = x; return; }
    int i = (x >> t) & 1;
    if(!ch[u][i]) ch[u][i] = ++sz;
    insert(ch[u][i], t - 1, x);
}

int query(int u, int t, int x){
    if(t < 0) return value[u];
    int i = (x >> t) & 1;
    if(ch[u][!i]) return query(ch[u][!i], t-1, x);
    return query(ch[u][i], t - 1, x);
}

void solve(){
    n = read(); m = read();
    memset(ch, 0, sizeof ch);
    sz = rt = 1;
    memset(value, 0, sizeof value);
    for(int i = 1; i <= n; i++) a[i] = read(), insert(rt, 31, a[i]);

    printf("Case #%d:\n", t - 1);
    for (int i = 1; i <= m; i++){
        int s = read();
        printf("%d\n", query(rt, 31, s));
    }
}

int main(){
    t = 1; T = read();
    while(t++ <= T) solve();
    return 0;
}

异或最小值

Vitya and Strange Lesson

$(a_1,a_2,a_3..a_n) \oplus A\oplus B$ 根据结合率，等价于 $(a_1,a_2,a_3..a_n) \oplus (A\oplus B)$ 。

因为我们要求的是 $mex$ ，那么我们就考虑 $0$ 到 $mex-1$ 的数是否全部出现过。那么就变为查询异或最小值了，我们在构建 $\text{01 Trie}$ 时要顺便记录 $\text{Trie}$ 中的元素个数，也就是$sz$，当一个节点的元素个数填满时，我们是不能考虑的。

/*
https://ac.nowcoder.com/acm/problem/112209
*/
#include<bits/stdc++.h>
using namespace std;

const int maxn = 6e6 + 10;
int read(){ int x; scanf("%d", &x); return x;}
int sz[maxn], ch[maxn][2];
int size = 1, rt = 1;
int n, m, last;

void insert(int u, int t, int x){
    if(t < 0){sz[u] = 1; return;}
    int i = (x >> t) & 1;
    if(!ch[u][i]) ch[u][i] = ++size;
    insert(ch[u][i], t - 1, x);
    sz[u] = sz[ch[u][i]] + sz[ch[u][!i]];
}
int query(int u, int t, int x){
    if(t < 0 || u == 0) return 0;
    int i = (x >> t) & 1;
    if((1 << t) != sz[ch[u][i]]) return query(ch[u][i], t - 1, x);
    else return query(ch[u][!i], t - 1, x) + (1 << t);
}

int main(){
    n = read(); m = read(); 
    for(int i = 1; i <= n; i++) insert(rt, 20, read());
    while(m--){
        last ^= read();
        printf("%d\n", query(rt, 20, last));
    }
    return 0;
}

保存值的异或最大值

奶牛异或

$a\oplus a = 0$ ，我们可以考虑做一次前缀异或和。那么区间操作就变为单点操作了。

/*
https://ac.nowcoder.com/acm/problem/22998
*/
#include<bits/stdc++.h>
using namespace std;

const int maxn = 2e7 + 100;
int ch[maxn][2], id[maxn], a[maxn];

int sz = 0, rt, n;
int ans1, ans2 = 1, ans3 = 1;

int read() {int x; scanf("%d", &x); return x;}

void insert(int u, int t, int x, int ID){
    if(t < 0){ id[u] = ID; return; }
    int f = ((x >> t) & 1);
    if(!ch[u][f]) ch[u][f] = ++sz;
    insert(ch[u][f], t-1, x, ID);
}

int ask(int u, int t, int x){
    if(t < 0) return id[u];
    int f = ((x >> t) & 1);
    if(ch[u][!f]) return ask(ch[u][!f], t-1, x);
    
    return ask(ch[u][f], t-1, x);
}

int main(){
    n = read(); a[0] = 0;
    for(int i = 1; i <= n; i++) (a[i] = a[i-1] ^ read());
    rt = ++ sz;
    for(int i = 1; i <= n; i++){
        insert(rt, 22, a[i-1], i);
        int x = ask(rt, 22, a[i]) - 1;
        if(ans1 < (a[x] ^ a[i])){
            ans1 = (a[x] ^ a[i]);
            ans2 = x + 1;
            ans3 = i;
        }
    }
    printf("%d %d %d\n", ans1, ans2, ans3);
    return 0;
}

带删除的异或最小值

Perfect Security

我们在$\text{01 Trie}$ 上再维护一个$sz_i$ 标记，表示是否这个节点下还有没有可用元素。那么删除和插入都只会影响一条链。

/*
https://ac.nowcoder.com/acm/problem/112567 
*/
#include<bits/stdc++.h>
using namespace std;

int read(){int x; scanf("%d", &x); return x;}

const int maxn = 6e6 + 10;
int ch[maxn][2], sz[maxn];
int size = 0, rt = 0;
int n, m;
int a[maxn], b[maxn], c[maxn];

void insert(int u, int t, int x){
    if(t < 0) return;
    int i = (x >> t) & 1;
    if(!ch[u][i])  ch[u][i] = ++size;;
    sz[ch[u][i]]++;
    insert(ch[u][i], t - 1, x);
}

void erase(int u, int t, int x){
    if(t < 0) return;
    int i = (x >> t) & 1;
    sz[ch[u][i]]--;
    erase(ch[u][i], t - 1, x);
}

int query(int u, int t, int x){
    if(t < 0) return 0;
    int i = (x >> t) & 1;
    if(sz[ch[u][i]]) return query(ch[u][i], t-1, x) + (i << t);
    else return query(ch[u][!i], t-1, x) + ((!i) << t);
}
int main(){
    n = read(); 
    for(int i = 1; i <= n; i++) a[i] = read();
    for(int i = 1; i <= n; i++) b[i] = read(), insert(rt, 30, b[i]);

    for(int i = 1; i <= n; i++){
        int x = query(rt, 30, a[i]);
        printf("%d\n", a[i] ^ x);
        erase(rt, 30, x);
    }
    return 0;
}

带删除的异或最大值

D. Vasiliy's Multiset

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

"+ x" — add integer x to multiset A.
"- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
"? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer$q (1 ≤ q≤ 200 000) $— the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer $x_i (1 ≤ x_i ≤ 10^9)$. It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer x**i and some integer from the multiset A.

input

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11

output

Note

After first five operations multiset A contains integers $0, 8, 9, 11, 6 and 1.$

The answer for the sixth query is integer — maximum among integers , , , and .

题意

初始有个一个空集，n个操作，操作分三种

＋x，将一个 x 加入集合
－x，删除集合内的一个 x
？x，询问集合中与 x 异或的最大值

代码

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

const int maxn = 6e6 + 10;
int ch[maxn][2], sz[maxn], val[maxn], cnt[maxn];
int size = 0, rt = 0;
int n, m;

void insert(int u, int t, int x){
    if(t < 0) { val[u] = x; return;}
    int i = (x >> t) & 1;
    if(!ch[u][i])  ch[u][i] = ++size;
    sz[ch[u][i]]++;
    insert(ch[u][i], t - 1, x);
}

void erase(int u, int t, int x){
    if(t < 0) return;
    int i = (x >> t) & 1;
    sz[ch[u][i]]--;
    erase(ch[u][i], t - 1, x);
}

int query(int u, int t, int x){
    if(t < 0) return val[u];
    int i = (x >> t) & 1;
    if(sz[ch[u][!i]]) return query(ch[u][!i], t-1, x);
    else return query(ch[u][i], t-1, x);
}

int main(){
    cin >> n;
    
    for(int i = 1; i <= n; i++){
        char c; cin >> c;
        int t; cin >> t;
        if(c == '+') insert(rt, 30, t);
        else if(c == '-') erase(rt, 30, t);
        else  printf("%d\n", max(t, query(rt, 30, t) ^ t));
    }
    return 0;
}
/*
12
? 1
+ 1
+ 4
? 5
? 6
*/

posted @ 2020-11-19 20:53 Frnas 阅读(182) 评论(0) 编辑收藏举报

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Frnas

日常codeforces，不过目前在休假，只能第二天起来打virtual。

01 Trie 专题

01 Trie 专题

异或最大值

The xor largest pair

Xor Sum

代码

异或最小值

Vitya and Strange Lesson

保存值的异或最大值

奶牛异或

带删除的异或最小值

带删除的异或最大值

D. Vasiliy's Multiset

题意

代码

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