题目链接：uva 1519 - Dictionary Size

题目大意：给出n个字符串组成的字典，现在要添加新的单词，从已有单词中选出非空前缀和非空后缀，组成新单词。问说能组成多少个单词。

解题思路：建立一棵前缀树和一棵后缀树，有多少节点即为有多少个前缀，扣除中间的部分即可加上长度为1的字符串即可。

#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];
    }
}

uva 1519 - Dictionary Size(字典树)