01 Trie 專題
01 Trie 專題
異或最大值
The xor largest pair
題意： 異或最大值的模板。一個數和一個序列中一個數的異或最大值是多少？要支援詢問。
思路:考慮把序列插入，構建一個 \(\text{Trie}\) 樹。那麼在詢問時，只需要討論該數的位是 \(0\) 還是 \(1\) 就行了。這樣就需要 \(O(n\log w)\) 的預處理，\(O(\log w)\) 的詢問和修改，為什麼是對的。因為異或中我們如果可以滿足最高位，那麼沒有理由不改變最高位，因為 \(2^{bit} >\sum_{i=0}^{i < bit}2^i\) 。那麼由高位到地位貪心就可以了。
/* 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 做了一個遊戲，Prometheus 給 Zeus 一個集合，集合中包含了N個正整數，隨後 Prometheus 將向 Zeus 發起M次詢問，每次詢問中包含一個正整數 S ，之後 Zeus 需要在集合當中找出一個正整數 K ，使得 K 與 S 的異或結果最大。Prometheus 為了讓 Zeus 看到人類的偉大，隨即同意 Zeus 可以向人類求助。你能證明人類的智慧麼？
Input
輸入包含若干組測試資料，每組測試資料包含若干行。
輸入的第一行是一個整數\(T（T < 10）\)，表示共有T組資料。
每組資料的第一行輸入兩個正整數\(N，M（<1=N,M<=100000）\)
Output
對於每組資料，首先需要輸出單獨一行”Case #?:”，其中問號處應填入當前的資料組數，組數從1開始計算。
對於每個詢問，輸出一個正整數K，使得K與S異或值最大。
Sample Input
2
3 2
3 4 5
1 5
4 1
4 6 5 6
3
Sample Output
Case #1:
4
3
Case #2:
4
同上題，簡單板子
程式碼
/* 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; }
異或最小值
Vitya and Strange Lesson
\((a_1,a_2,a_3..a_n) \oplus A\oplus B\) 根據結合率，等價於 \((a_1,a_2,a_3..a_n) \oplus (A\oplus B)\) 。
因為我們要求的是 \(mex\) ，那麼我們就考慮 \(0\) 到 \(mex-1\) 的數是否全部出現過。那麼就變為查詢異或最小值了，我們在構建 \(\text{01 Trie}\) 時要順便記錄 \(\text{Trie}\) 中的元素個數，也就是\(sz\)，當一個節點的元素個數填滿時，我們是不能考慮的。
/*
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;
}
儲存值的異或最大值
奶牛異或
\(a\oplus a = 0\) ，我們可以考慮做一次字首異或和。那麼區間操作就變為單點操作了。
/*
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;
}
帶刪除的異或最小值
我們在\(\text{01 Trie}\) 上再維護一個\(sz_i\) 標記，表示是否這個節點下還有沒有可用元素。那麼刪除和插入都只會影響一條鏈。
/*
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;
}
帶刪除的異或最大值
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
11
10
14
13
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 .
題意
初始有個一個空集，n個操作，操作分三種
- ＋x，將一個 x 加入集合
- －x，刪除集合內的一個 x
- ？x，詢問集合中與 x 異或的最大值
程式碼
#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
*/