字典序第一(First) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <algorithm>
#include <iostream>
#include <queue>
#include <string>
#include <vector>
using namespace std;

struct Node {
    int ch[26];
    bool is_end;
    Node() {
        fill(begin(ch), end(ch), 0);
        is_end = false;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<string> words(n);
    vector<Node> tr;
    tr.reserve(300005);
    tr.emplace_back();
    for (int i = 0; i < n; ++i) {
        cin >> words[i];
        int u = 0;
        for (char ch : words[i]) {
            int c = ch - 'a';
            if (!tr[u].ch[c]) {
                tr[u].ch[c] = (int)tr.size();
                tr.emplace_back();
            }
            u = tr[u].ch[c];
        }
        tr[u].is_end = true;
    }

    vector<string> ans;
    for (int i = 0; i < n; ++i) {
        const string& s = words[i];
        int u = 0;
        bool prefix_conflict = false;
        int adj[26] = {0};
        for (char ch : s) {
            if (tr[u].is_end) {
                prefix_conflict = true;
                break;
            }
            int c = ch - 'a';
            for (int o = 0; o < 26; ++o) {
                if (o != c && tr[u].ch[o]) {
                    adj[c] |= (1 << o);
                }
            }
            u = tr[u].ch[c];
        }
        if (prefix_conflict) continue;

        int indeg[26] = {0};
        for (int a = 0; a < 26; ++a) {
            for (int b = 0; b < 26; ++b) {
                if ((adj[a] >> b) & 1) indeg[b]++;
            }
        }
        queue<int> q;
        for (int a = 0; a < 26; ++a) {
            if (indeg[a] == 0) q.push(a);
        }
        int visited = 0;
        while (!q.empty()) {
            int x = q.front();
            q.pop();
            visited++;
            for (int y = 0; y < 26; ++y) {
                if ((adj[x] >> y) & 1) {
                    if (--indeg[y] == 0) q.push(y);
                }
            }
        }
        if (visited == 26) ans.push_back(s);
    }

    cout << ans.size() << "\n";
    for (const auto& s : ans) cout << s << "\n";
    return 0;
}