Problem Description
Ignatius is so lucky that he met a Martian yesterday. But he didn't know the language the Martians use. The Martian gives him a history book of Mars and a dictionary when it leaves. Now Ignatius want to translate the history book into English. Can you help him?
Input
The problem has only one test case, the test case consists of two parts, the dictionary part and the book part. The dictionary part starts with a single line contains a string "START", this string should be ignored, then some lines follow, each line contains two strings, the first one is a word in English, the second one is the corresponding word in Martian's language. A line with a single string "END" indicates the end of the directory part, and this string should be ignored. The book part starts with a single line contains a string "START", this string should be ignored, then an article written in Martian's language. You should translate the article into English with the dictionary. If you find the word in the dictionary you should translate it and write the new word into your translation, if you can't find the word in the dictionary you do not have to translate it, and just copy the old word to your translation. Space(' '), tab(' '), enter('
') and all the punctuation should not be translated. A line with a single string "END" indicates the end of the book part, and that's also the end of the input. All the words are in the lowercase, and each word will contain at most 10 characters, and each line will contain at most 3000 characters.
Output
In this problem, you have to output the translation of the history book.
题意：这题就是翻译文章，所以把所有译文存到数组中，对每个单词插入进字典树，并在结点上增加译文在数组中位置的信息，输出时只需查找单词，若存在则输出译文，否组输出原文即可。
依旧是简单的字典树模版题。
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
using namespace std;
char s[30] , sl[505000][30] , sg[30];
struct TnT {
TnT *next[27];
int v;
TnT():v(0) {
memset(next , 0 , sizeof(next));
}
};
void build(TnT *root , char s[] , int num) {
int len = strlen(s);
TnT *p = root;
for(int i = 0 ; i < len ; i++) {
int id = s[i] - 'a';
if(p->next[id] == NULL) {
p->next[id] = new TnT;
}
p = p->next[id];
}
p->v = num;
}
int find(TnT *root , char s[]) {
int temp = 0;
int len = strlen(s);
TnT *p = root;
for(int i = 0 ; i < len ; i++) {
int id = s[i] - 'a';
if(p->next[id] == NULL) {
return 0;
}
p = p->next[id];
}
return p->v;
}
void de(TnT *root) {
if(root == NULL) {
return;
}
else {
for(int i = 0 ; i < 26 ; i++) {
de(root->next[i]);
}
}
delete root;
}
int main()
{
TnT *root = new TnT;
int count = 1;
while(gets(s)) {
if(strcmp(s , "START") == 0) {
continue;
}
if(strcmp(s , "END") == 0) {
break;
}
int len = strlen(s);
int flag = 0;
int temp = 0;
for(int i = 0 ; i < len ; i++) {
if(flag == 0 && s[i] >= 'a' && s[i] <= 'z') {
sl[count][temp++] = s[i];
}
if(flag == 1 && s[i] >= 'a' && s[i] <= 'z') {
sg[temp++] = s[i];
}
if(s[i] == ' ') {
flag = 1;
temp = 0;
}
}
build(root , sg , count);
count++;
memset(sg , 0 , sizeof(sg));
}
while(gets(s)) {
if(strcmp(s , "START") == 0) {
continue;
}
if(strcmp(s , "END") == 0) {
break;
}
int len = strlen(s);
int temp = 0;
for(int i = 0 ; i < len ; i++) {
if(s[i] >= 'a' && s[i] <= 'z') {
sg[temp++] = s[i];
}
else {
temp = 0;
int gg = find(root , sg);
if(gg) {
printf("%s" , sl[gg]);
}
else {
printf("%s" , sg);
}
memset(sg , 0 , sizeof(sg));
printf("%c" , s[i]);
}
}
printf("
");
}
return 0;
}