学步园

题目链接：poj 2503 Babelfish

题目大意：给定若干个字符串间的映射关系，然后是若干次查询，每次查询一个字符串，判断该字符串是否有映射串，

有的话输出映射串，否则输出eh。

解题思路：字典树水题。

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

using namespace std;

const int maxn = 1e6+5;
const int maxm = 105;
const int sigma_size = 26;

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;
char w[100005][maxm], op[maxm], s[maxm];

int main () {

    T.init();
    strcpy(w[0], "eh");
    while (gets(op) && strcmp(op, "")) {
        sscanf(op, "%s%s", w[++N], s);
        T.insert(s, N);
    }

    while (gets(op)) {
        int x = T.find(op);
        printf("%s\n", w[x]);
    }
    return 0;
}

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

int Tire::idx (char ch) {
    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];
    }
    val[u] = x;;
}