Ultraviolet A 1462-fuzzy Google suggest (Dictionary tree + DFS)

Link to the question: Ultraviolet A 1462-fuzzy Google suggest

Simulate a fuzzy search by Google. given to a string set, there are then n searches. Each time there is an integer x and a string, which indicates that the string can be modified X times, this includes adding, modifying, and deleting a character. The modified character may be the prefix of the number of strings in the character set.

Solution: first, create a dictionary tree. For each search, perform DFS on the dictionary tree and process the data according to the position where the parameter x matches the string. Then, mark the position matching to the end as 2, then, for the second DFS, search for the first occurrence location of 2 on each branch.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 3000005;const int sigma_size = 26;struct Tire {    int sz;    int g[maxn][sigma_size];    int val[maxn], vis[maxn];    int deep, ans;    char s[15];    void init();    int idx(char ch);    void insert(char* str);    int solve(char* str, int x);    void dfs(int u, int d, int x);    void clear(int u, int flag);}AC;int main () {    int n, x;    char str[15];    while (scanf("%d", &n) == 1) {        AC.init();        for (int i = 0; i < n; i++) {            scanf("%s", str);            AC.insert(str);        }        scanf("%d", &n);        for (int i = 0; i < n; i++) {            scanf("%s%d", str, &x);            printf("%d\n", AC.solve(str, x));        }    }    return 0;}void Tire::init() {    sz = 1;    val[0] = 0;    //memset(vis, 0, sizeof(vis));    memset(g[0], 0, sizeof(g[0]));}int Tire::idx (char ch) {    return ch - ‘a‘;}void Tire::dfs (int u, int d, int x) {    if (x < 0)        return;    if (vis[u] == 0)        vis[u] = 1;    if (d == deep) {        vis[u] = 2;        return;    }    int v = idx(s[d]);    if (g[u][v])        dfs(g[u][v], d + 1, x);    dfs(u, d + 1, x - 1);    for (int i = 0; i < sigma_size; i++) {        if (g[u][i]) {            dfs(g[u][i], d, x - 1);            dfs(g[u][i], d + 1, x - 1);        }    }}void Tire::clear(int u, int flag) {    if (vis[u] == 0)        return;    if (flag && vis[u] == 2) {        ans += val[u];        flag = 0;    }    for (int i = 0; i < sigma_size; i++) {        if (g[u][i])            clear(g[u][i], flag);    }    vis[u] = 0;}int Tire::solve(char* str, int x) {    ans = 0;    deep = strlen(str);    strcpy(s, str);    dfs(0, 0, x);    clear(0, 1);    return ans;}void Tire::insert(char* str) {    int u = 0, n = strlen(str);    for (int i = 0; i < n; i++) {        int v = idx(str[i]);        if (g[u][v] == 0) {            val[sz] = 0;            memset(g[sz], 0, sizeof(g[sz]));            g[u][v] = sz++;        }        u = g[u][v];        val[u]++;    }}

Ultraviolet A 1462-fuzzy Google suggest (Dictionary tree + DFS)