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

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

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

    string str[];

    void solve() {
        int n;
        cin>>n;
        clear();
        for(int i=;i<=n;i++) {
            cin>>str[i];
            for(int j=;j<str[i].length();j++) str[i][j] -= '';
            insert(str[i]);
        }
        int flag=;
        for(int i=;i<=n;i++) {
            if(query(str[i])) {
                flag=;
                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 ;
}

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

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

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

string str[];

bool solve(int t) {
    clear();
    int n = ;
    while (cin >> str[++n]) {
        if (str[n][] == '')
            break;
        for (int j = ; j < str[n].length(); j++) str[n][j] -= '';
        insert(str[n]);
    }
    --n;
    if (n <= )
        return false;
    int flag = ;
    for (int i = ; i <= n; i++) {
        if (query(str[i])) {
            flag = ;
            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 = ;
    ios::sync_with_stdio(false);
    while (Trie::solve(++t))
        ;
    return ;
}

#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[];

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

int __cnt = ;

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

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

void clear() {
    ind = ;
    root = newnode();
}
void insert(string &s) {
    Node *pos = root;
    for (int i = ; 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 + ] = , ans = max(ans, i + );
    }
}

string str[];
string art;

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

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

#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[];
    int val, sum;
    Node *clear() {
        for (int i = ; i < ; i++) ch[i] = NULL;
        val = sum = ;
        return this;
    }
};

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

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

string str[];

void solve() {
    int m, n;
    cin >> m >> n;
    clear();
    for (int i = ; i <= m; i++) {
        int len;
        cin >> len;
        insert(len);
    }
    dfs(root);
    for (int i = ; 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 ;
}

#10056. 「一本通 2.3 练习 5」The XOR-longest Path

树上前缀异或和以后直接转化为两点最大异或和。其实用欧拉序也可以。

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

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

void dfs(int p) {
    vis[p]=;
    for(int i=;i<g[p].size();i++) {
        int q=g[p][i].first;
        if(vis[q]==) {
            Trie::a[q]=Trie::a[p] ^ g[p][i].second;
            dfs(q);
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=;i<n;i++) {
        int u,v,w;
        cin>>u>>v>>w;
        g[u].push_back(make_pair(v,w));
        g[v].push_back(make_pair(u,w));
    }
    dfs();
    Trie::solve(n);
}