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
, 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;
}