Description

Luogu3065

Solution

首先，一個串要是最小的，別的串不能是它的字首，且和它有相同字首的串字典序都比他小。
Trie樹是顯然要用的，難點在於如何判斷能否最小。其實這之中蘊含了字母間的大小關係，即同一層中，欽定的最小串的字元要比其它的大。然後就可以建一個有向圖跑拓撲排序判環就行。

Code

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <string>
#include <vector>

namespace wyx {

typedef int ll;

ll read() {
    ll ans = 0, fl = 1;
    char c;
    for (c = getchar(); c > '9' || c < '0'; c = getchar())
        if (c == '-') fl = -1;
    ans = c - '0';
    for (c = getchar(); c >= '0' && c <= '9'; c = getchar())
        ans = ans * 10 + c - '0';
    return fl * ans;
}

typedef double ld;
typedef unsigned long long ull;
const int N = 30010;

struct Trie {
    int to[N * 10][26], cnt, val[N * 10];
    Trie() {
        memset(to, 0, sizeof to);
        memset(val, 0, sizeof val);
        cnt = 0;
    }
    void add(const std::string& s) {
        int len = s.length();
        int now = 0;
        for (int i = 0; i < len; ++i) {
            if (!to[now][s[i] - 'a']) to[now][s[i] - 'a'] = ++cnt;
            now = to[now][s[i] - 'a'];
        }
        val[now]++;
    }
} Tr;
struct Graph {
    std::vector<int> g[26];
    int deg[26];
    void init() {
        for (int i = 0; i < 26; ++i) g[i].clear(), deg[i] = 0;
    }
    void adde(int x, int y) {
        g[x].push_back(y);
        deg[y]++;
    }
    bool toposort() {
        std::queue<int> q;
        for (int i = 0; i < 26; ++i)
            if (deg[i] == 0) q.push(i);
        while (!q.empty()) {
            int x = q.front();
            q.pop();
            for (int i : g[x])
                if (--deg[i] == 0) q.push(i);
        }
        for (int i = 0; i < 26; ++i)
            if (deg[i]) return false;
        return true;
    }
} G;
std::string s[N];
int n, vis[N];

bool work(int x) {
    int now = 0, len = s[x].length();
    for (int i = 0; i < len; ++i) {
        if (Tr.val[now]) return false;
        for (int j = 0; j < 26; ++j) {
            if (j != s[x][i] - 'a' && Tr.to[now][j]) G.adde(s[x][i] - 'a', j);
        }
        now = Tr.to[now][s[x][i] - 'a'];
    }
    return true;
}

void main() {
    n = read();
    for (int i = 1; i <= n; ++i) std::cin >> s[i], Tr.add(s[i]);
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
        G.init();
        if (work(i) && G.toposort()) {
            vis[i] = 1;
            cnt++;
        }
    }
    printf("%d\n", cnt);
    for (int i = 1; i <= n; ++i)
        if (vis[i]) {
            std::cout << s[i] << std::endl;
        }
}

};  // namespace wyx

int main() {
    wyx::main();
    return 0;
}