Babelfish+poj+字典树

Babelfish

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 32392 Accepted: 13929

Description

You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

Input

Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

Output

Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".

Sample Input

dog ogdaycat atcaypig igpayfroot ootfrayloops oopslayatcayittenkayoopslay

Sample Output

catehloops

解决方案：该此题用上字典树效率蛮高的，我用的是孩子兄弟建树法，head指向孩子节点，next指向兄弟节点。w了几次，发现自己定义的最大节点个数少了，最大节点个数应为(单词个数)*(最长单词长度)；

code:
#include<iostream>#include<cstdio>#include<cstring>#define MMAX  1000003using namespace std;int head[MMAX];///指向左孩子int next[MMAX];///指向兄弟节点char word[MMAX];int sz;char ke[MMAX][15];int val[MMAX];void Insert(char *w,char *key){    int u=0,v;    int len=strlen(w);    for(int i=0; i<len; i++)    {        for(v=head[u]; v; v=next[v])        {            if(word[v]==w[i]) break;        }        if(!v)        {            v=sz++;            next[v]=head[u];            head[u]=v;            head[v]=0;            val[v]=0;            word[v]=w[i];        }        u=v;    }    val[v]=1;    strcpy(ke[v],key);}void query(char *Find){    int len=strlen(Find);    int u=0,v;    bool flag=true;    for(int i=0; i<len; i++)    {        for(v=head[u]; v; v=next[v])        {            if(Find[i]==word[v]) break;        }        if(!v)        {            flag=false;            break;        }        u=v;    }    if(!flag||!val[v])    {        printf("eh\n");    }    else    {        printf("%s\n",ke[v]);    }}int main(){    char word[40],kk1[15],kk[15];    memset(head,0,sizeof(head));    memset(next,0,sizeof(next));    memset(val,0,sizeof(val));    sz=1;    while(gets(word)&&word[0]!='\0')    {        sscanf(word,"%s %s",kk1,kk);        Insert(kk,kk1);    }    while(gets(kk)&&kk[0]!='\0')    {        query(kk);    }    return 0;}