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[一本通学习笔记] 字典树与 0-1 Trie

字典树中根到每个结点对应原串集合的一个前缀，这个前缀由路径上所有转移边对应的字母构成。我们可以对每个结点维护一些需要的信息，这样即可以去做很多事情。

#10049. 「一本通 2.3 例 1」Phone List

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

namespace Trie {
    struct Node {
        Node *ch[10];
        int val;
        Node* clear() {
            for(int i=0;i<10;i++) ch[i]=NULL;
            val=0;
            return this;
        }
    };

    Node *root;
    Node pool[1000005];
    int ind=0;
    Node* newnode() {
        return pool[ind++].clear();
    }

    void clear() {
        ind=0;
        root=newnode();
    }
    void insert(string s) {
        Node *pos=root;
        for(int i=0;i<s.length();i++) {
            if(pos->ch[s[i]]==NULL) pos->ch[s[i]]=newnode();
            pos->val++;
            pos=pos->ch[s[i]];
        }
    }
    int query(string s) {
        Node *pos=root;
        for(int i=0;i<s.length();i++) {
            if(pos->ch[s[i]]==NULL) return 0;
            pos=pos->ch[s[i]];
        }
        return pos->val;
    }

    string str[10005];

    void solve() {
        int n;
        cin>>n;
        clear();
        for(int i=1;i<=n;i++) {
            cin>>str[i];
            for(int j=0;j<str[i].length();j++) str[i][j] -= '0';
            insert(str[i]);
        }
        int flag=0;
        for(int i=1;i<=n;i++) {
            if(query(str[i])) {
                flag=1;
                break;
            }
        }
        if(flag) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
}

int main() {
    int t;
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--) Trie::solve();
    return 0;
}

#10050. 「一本通 2.3 例 2」The XOR Largest Pair

0-1 Trie通常用于异或相关的问题，思路是记录所有01串后，我们找最大异或和的时候可以从高位到低位贪心，此时Trie发挥的作用就是在当前前缀已经选择的情况下，能使得正在考虑位异或为1的后缀是否存在。

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

namespace Trie {
struct Node {
    Node *ch[2];
    Node *clear() {
        ch[0] = ch[1] = 0;
        return this;  // Don't forget this
    }
};
Node *root;
Node pool[4000005];
int ind;
Node *newnode() { return pool[ind++].clear(); }
void insert(int x) {
    Node *pos = root;
    for (int i = 30; i >= 0; --i) {
        int b = (x / (1 << i)) & 1;
        if (pos->ch[b] == NULL)
            pos->ch[b] = newnode();
        pos = pos->ch[b];
    }
}
int query(int x) {
    Node *pos = root;
    int ans = 0;
    for (int i = 30; i >= 0 && pos != NULL; --i) {
        int b = (x / (1 << i)) & 1;
        if (pos->ch[b ^ 1] != NULL)
            pos = pos->ch[b ^ 1], ans += (1 << i);
        else
            pos = pos->ch[b];
    }
    return ans;
}
int a[1000005];
void solve() {
    int n, ans = 0;
    cin >> n;
    root = newnode();  // Don't forget this
    for (int i = 1; i <= n; i++) cin >> a[i], insert(a[i]);
    for (int i = 1; i <= n; i++) ans = max(ans, query(a[i]));
    cout << ans << endl;
}
}  // namespace Trie

int main() {
    ios::sync_with_stdio(false);
    Trie::solve();
}

#10051. 「一本通 2.3 例 3」Nikitosh 和异或

看到这个算式我们很容易想到前缀和转化。问题转化为求s[r1]^s[l1-1] + s[r2]^s[l2-1]最大。

由于要r1<l2，所以可考虑去处理出一个前缀max和一个后缀max，然后枚举分界点。那么答案就是

Max{pre[i]+suf[i+1]}

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

namespace Trie {
struct Node {
    Node *ch[2];
    Node *clear() {
        ch[0] = ch[1] = 0;
        return this;  // Don't forget this
    }
};
Node *root;
Node pool[8000005];
int ind;
Node *newnode() { return pool[ind++].clear(); }
void insert(int x) {
    Node *pos = root;
    for (int i = 30; i >= 0; --i) {
        int b = (x / (1 << i)) & 1;
        if (pos->ch[b] == NULL)
            pos->ch[b] = newnode();
        pos = pos->ch[b];
    }
}
int query(int x) {
    Node *pos = root;
    int ans = 0;
    for (int i = 30; i >= 0 && pos != NULL; --i) {
        int b = (x / (1 << i)) & 1;
        if (pos->ch[b ^ 1] != NULL)
            pos = pos->ch[b ^ 1], ans += (1 << i);
        else
            pos = pos->ch[b];
    }
    return ans;
}
int a[1000005], pre[1000005], suf[1000005];
void solve() {
    int n, ans = 0;
    cin >> n;
    root = newnode();  // Don't forget this
    for (int i = 1; i <= n; i++) cin >> a[i], insert(a[i]), pre[i] = max(pre[i - 1], query(a[i]));
    ind = 0;
    root = newnode();
    reverse(a + 1, a + n + 1);
    for (int i = 1; i <= n; i++) insert(a[i]), suf[i] = max(suf[i - 1], query(a[i]));
    reverse(suf + 1, suf + n + 1);
    for (int i = 1; i < n; i++) ans = max(ans, pre[i] + suf[i + 1]);
    cout << ans << endl;
}
}  // namespace Trie

int main() {
    ios::sync_with_stdio(false);
    Trie::solve();
}

#10052. 「一本通 2.3 练习 1」Immediate Decodability

和前面那题一样

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

namespace Trie {
struct Node {
    Node *ch[2];
    int val;
    Node *clear() {
        for (int i = 0; i < 2; i++) ch[i] = NULL;
        val = 0;
        return this;
    }
};

Node *root;
Node pool[1000005];
int ind = 0;
Node *newnode() { return pool[ind++].clear(); }

void clear() {
    ind = 0;
    root = newnode();
}
void insert(string s) {
    Node *pos = root;
    for (int i = 0; i < s.length(); i++) {
        if (pos->ch[s[i]] == NULL)
            pos->ch[s[i]] = newnode();
        pos->val++;
        pos = pos->ch[s[i]];
    }
}
int query(string s) {
    Node *pos = root;
    for (int i = 0; i < s.length(); i++) {
        if (pos->ch[s[i]] == NULL)
            return 0;
        pos = pos->ch[s[i]];
    }
    return pos->val;
}

string str[10005];

bool solve(int t) {
    clear();
    int n = 0;
    while (cin >> str[++n]) {
        if (str[n][0] == '9')
            break;
        for (int j = 0; j < str[n].length(); j++) str[n][j] -= '0';
        insert(str[n]);
    }
    --n;
    if (n <= 0)
        return false;
    int flag = 0;
    for (int i = 1; i <= n; i++) {
        if (query(str[i])) {
            flag = 1;
            break;
        }
    }
    if (flag)
        cout << "Set " << t << " is not immediately decodable" << endl;
    else
        cout << "Set " << t << " is immediately decodable" << endl;
    return true;
}
}  // namespace Trie

int main() {
    int t = 0;
    ios::sync_with_stdio(false);
    while (Trie::solve(++t))
        ;
    return 0;
}

#10053. 「一本通 2.3 练习 2」L 语言

我们记录u[i]表示文章的每一个前缀s[1..i]是否可被理解。做一个类似dp的处理即可。

刚开始忘记传引用T了半天……

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

char buf[1000005];

void readstr(string &tar) {
    scanf("%s", buf);
    tar = buf;
}

int __cnt = 0;

namespace Trie {
struct Node {
    Node *ch[26];
    int val;
    Node *clear() {
        for (int i = 0; i < 26; i++) ch[i] = NULL;
        val = 0;
        return this;
    }
};

Node *root;
Node pool[5000005];
int u[1000005];
int ind = 0, ans = 0;
Node *newnode() { return pool[ind++].clear(); }

void clear() {
    ind = 0;
    root = newnode();
}
void insert(string &s) {
    Node *pos = root;
    for (int i = 0; i < s.length(); i++) {
        if (pos->ch[s[i]] == NULL)
            pos->ch[s[i]] = newnode();
        pos = pos->ch[s[i]];
    }
    pos->val++;
}
void query(string &s, int start) {
    Node *pos = root;
    int len = s.length();
    for (int i = start; i < len; i++) {
        __cnt++;
        if (pos->ch[s[i]] == NULL)
            return;
        pos = pos->ch[s[i]];
        if (pos->val)
            u[i + 1] = 1, ans = max(ans, i + 1);
    }
}

string str[1005];
string art;

void solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    clear();
    for (int i = 1; i <= n; i++) {
        readstr(str[i]);
        for (int j = 0; j < str[i].length(); j++) str[i][j] -= 'a';
        insert(str[i]);
    }
    for (int i = 1; i <= m; i++) {
        ans = 0;
        memset(u, 0, sizeof u);
        readstr(art);
        int len = art.length();
        for (int j = 0; j < len; j++) art[j] -= 'a';
        u[0] = 1;
        for (int j = 0; j < len; j++) {
            if (u[j] == 0)
                continue;
            query(art, j);
        }
        cout << ans << endl;
    }
}
}  // namespace Trie

int main() {
    int t;
    ios::sync_with_stdio(false);
    Trie::solve();
    // cout<<__cnt<<endl;
    return 0;
}

#10054. 「一本通 2.3 练习 3」Secret Message 秘密信息

对信息建Trie，仍然是在每个串的结束点上打标记。结果就等于把密码串丢上去跑，跑的路径上的标记和，加上最终停在的结点（如果整个密码串都成功匹配）的子树的标记和。前一个直接记录，后一个用树上前缀和预处理一下即可。

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

namespace Trie {
struct Node {
    Node *ch[2];
    int val, sum;
    Node *clear() {
        for (int i = 0; i < 2; i++) ch[i] = NULL;
        val = sum = 0;
        return this;
    }
};

Node *root;
Node pool[1000005];
int ind = 0;
Node *newnode() { return pool[ind++].clear(); }

void clear() {
    ind = 0;
    root = newnode();
}
void insert(int len) {
    Node *pos = root;
    for (int i = 0; i < len; i++) {
        int si;
        cin >> si;
        if (pos->ch[si] == NULL)
            pos->ch[si] = newnode();
        pos = pos->ch[si];
    }
    pos->val++;
}
void dfs(Node *p) {
    if (p == NULL)
        return;
    dfs(p->ch[0]);
    dfs(p->ch[1]);
    p->sum = p->val;
    if (p->ch[0])
        p->sum += p->ch[0]->sum;
    if (p->ch[1])
        p->sum += p->ch[1]->sum;
}
int query(int len) {
    Node *pos = root;
    int ans = 0;
    for (int i = 0; i < len; i++) {
        int si;
        cin >> si;
        ans += pos->val;
        if (pos->ch[si] == NULL) {
            for (int j = 1; j <= len - i - 1; j++) cin >> si;
            return ans;
        }
        pos = pos->ch[si];
    }
    return ans + pos->sum;
}

string str[10005];

void solve() {
    int m, n;
    cin >> m >> n;
    clear();
    for (int i = 1; i <= m; i++) {
        int len;
        cin >> len;
        insert(len);
    }
    dfs(root);
    for (int i = 1; i <= n; i++) {
        int len;
        cin >> len;
        cout << query(len) << endl;
    }
}
}  // namespace Trie

int main() {
    int t;
    ios::sync_with_stdio(false);
    Trie::solve();
    return 0;
}
01-16 05:41
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