poj 1816 Wild Words(字典树+DFS)

题目链接：poj 1816 Wild Words

题目大意：给定N个模板，即正则表达式，然后每次有一个询问串，输出能被哪些模板匹配。

解题思路：对模板建立字典树，然后每次询问即在字典树做DFS搜索，注意'*'的情况，可以匹配一个和多个，所以在结

尾的时候要注意。并且，模板串有重复的情况。

#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>

using namespace std;

const int maxn = 100000 * 6 + 5;
const int sigma_size = 28;

struct Tire {
    int sz;
    int g[maxn][sigma_size];
    int val[maxn];

    void init();
    int idx(char ch);
    void insert(char* s, int x);
    int find(char* s);
}T;

int N, M, L, F[maxn];
char W[30];
set<int> ans;

void dfs (int u, int p) {

    if (T.val[u] && p == L) {
        int k = T.val[u];
        while (k) {
            ans.insert(k - 1);
            k = F[k];
        }
    }


    if (T.g[u][27]) {
        for (int i = p; i <= L; i++)
            dfs(T.g[u][27], i);
    }

    if (p == L)
        return;

    if (T.g[u][26])
        dfs(T.g[u][26], p + 1);

    int v = W[p] - 'a';
    if (T.g[u][v])
        dfs(T.g[u][v], p + 1);
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        T.init();
        for (int i = 1; i <= N; i++) {
            scanf("%s", W);
            T.insert(W, i);
        }

        for (int i = 1; i <= M; i++) {
            scanf("%s", W);
            L = strlen(W);
            ans.clear();
            dfs(0, 0);

            if (ans.size()) {
                set<int>::iterator iter = ans.begin();
                printf("%d", *iter);
                for (iter++; iter != ans.end(); iter++)
                    printf(" %d", *iter);
                printf("\n");
            } else
                printf("Not match\n");
        }
    }
    return 0;
}

void Tire::init() {
    sz = 1;
    val[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Tire::idx (char ch) {
    if (ch == '?')
        return 26;
    if (ch == '*')
        return 27;
    return ch - 'a';
}

int Tire::find(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)
            return 0;
        u = g[u][v];
    }
    return val[u];
}

void Tire::insert(char* s, int x) {
    int u = 0, n = strlen(s);

    for (int i = 0; i < n; i++) {
        int v = idx(s[i]);

        if (g[u][v] == 0) {
            val[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }

        u = g[u][v];
    }
    F[x] = val[u];
    val[u] = x;
}