hdu 1075 What Are You Talking About(字典树)

题目链接：hdu 1075 What Are You Talking About

题目大意：给定若干个字符串的映射，要求输出给定字符串的原串，即可以用映射串代替的均要替换。

解题思路：对被映射串建立字典树，然后单词节点记录的是对应串映射串位置。然后对于需要翻译的字符串，逐个处理

处单词，在字典树中查找，如果有遍历到单词节点，则输出对应的映射串。

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

using namespace std;

const int maxn = 5000000;
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);
}S;

char r[500000][15], op[3005], word[15];

void solve (int& p) {
    if (p == 0)
        return;

    word[p++] = '\0';
    int x = S.find(word);
    if (x)
        printf("%s", r[x]);
    else
        printf("%s", word);
    p = 0;
}

int main () {
    while (scanf("%s", op) == 1) {
        S.init();
        int n = 1;

        while (scanf("%s", op), strcmp(op, "END")) {
            strcpy(r[n], op);
            scanf("%s", op);
            S.insert(op, n++);
        }

        scanf("%s", op);
        getchar();
        while (gets(op), strcmp(op, "END")) {
            int len = strlen(op), p = 0;
            for (int i = 0; i < len; i++) {
                if (op[i] >= 'a' && op[i] <= 'z')
                    word[p++] = op[i];
                else {
                    solve(p);
                    printf("%c", op[i]);
                }
            }
            solve(p);
            printf("\n");
        }
    }
    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;
}