Ultraviolet A 1519-dictionary size (Dictionary tree)

Link: Ultraviolet A 1519-dictionary size

A dictionary composed of N strings. Now you need to add new words and select non-empty prefixes and non-empty suffixes from existing words to form a new word. Ask how many words can be made up.

Solution: Create a Prefix Tree and a suffix tree, and add a string with a length of 1 after deducting the middle part.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 400005;const int sigma_size = 26;typedef long long ll;struct Tire {    int sz, g[maxn][sigma_size];    int c[sigma_size];    void init ();    int idx(char ch);    void insert(char* s);}pre, suf;int main () {    int N;    while (scanf("%d", &N) == 1 && N) {        char word[45];        int vis[sigma_size];        memset(vis, 0, sizeof(vis));        pre.init();        suf.init();        for (int i = 0; i < N; i++) {            scanf("%s", word);            int n = strlen(word);            if (n == 1)                vis[word[0] - ‘a‘] = 1;            pre.insert(word);            reverse(word, word + n);            suf.insert(word);        }        ll ans = (ll)(pre.sz - 1) * (suf.sz - 1);        for (int i = 0; i < sigma_size; i++)            ans -= (ll)pre.c[i] * suf.c[i] - vis[i];        printf("%lld\n", ans);    }    return 0;}void Tire::init () {    sz = 1;    memset(g[0], 0, sizeof(g[0]));    memset(c, 0, sizeof(c));}int Tire::idx (char ch) {    return ch - ‘a‘;}void Tire::insert(char* s) {    int u = 0, n = strlen(s);    for (int i = 0; i < n; i++) {        int v = idx(s[i]);        if (g[u][v] == 0) {            memset(g[sz], 0, sizeof(g[sz]));            g[u][v] = sz++;            if (i)                c[v]++;        }        u = g[u][v];    }}

Ultraviolet A 1519-dictionary size (Dictionary tree)