poj 2503 雜湊 Map 字典樹
阿新 • • 發佈:2019-02-02
Babelfish
Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".
Time Limit: 3000MS | Memory Limit: 65536K |
Total Submissions: 36967 | Accepted: 15749 |
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
Sample Input
dog ogday cat atcay pig igpay froot ootfray loops oopslay atcay ittenkay oopslay
Sample Output
cat eh loops
Hint
Huge input and output,scanf and printf are recommended.Source
//16096K 2625MS #include<iostream> #include<cstdio> #include<map> #include<cstring> using namespace std; int main() { char s[100],s1[11]; string ss; char c; map<string,string>Q; int num; while(gets(s)&&s[0]!='\0') //讀串比讀多個字元快 { int len=strlen(s); int i; for( i=0;i<len;i++) { if(s[i]==' ') {s[i]='\0'; break; } } ss=s+i+1; Q[ss]=s; } while(~scanf("%s",s1)) { if(Q[s1].size()) cout<<Q[s1]<<endl; else printf("eh\n"); } } // while(~scanf("%c",&c)) // 超時 // { // if(c=='\n') // break; // num=0; // while(c!=' ') // { // s[num++]=c; // scanf("%c",&c); // } // s[num]='\0'; //夠成字串 // num=0; // scanf("%c",&c); //防止上一個空格被讀入 // // while(c!='\n') // { // ss[num++]=c; // scanf("%c",&c); // } // ss[num]='\0'; // Q[ss]=s; // }
//字典樹 26240K <span id="transmark"></span>735MS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
char ss[100],s[100],c[100010][100];
int num=0;
struct node
{
int flag;
node *next[26];
}*head;
node * Creat()
{
node *p;
p=new node;
p->flag=0;
for(int i=0;i<26;i++)
p->next[i]=NULL;
return p;
}
int Build_Tree()
{
node *p=head;
int len=strlen(s);
for(int i=0;i<len;i++)
{
int a=s[i]-'a';
if(!p->next[a])
{
p->next[a]=Creat();
//p->next[a]->flag=num;
}
p=p->next[a];
}
p->flag=num;
}
int Find(char s1[])
{
int len=strlen(s1);
node *p=head;
for(int i=0;i<len;i++)
{
int a=s1[i]-'a';
if(!p->next[a])
{
return 0;
}
p=p->next[a];
}
return p->flag;
}
int main()
{
num=1;
head=Creat();
while(gets(ss)&&ss[0]!='\0')
{
sscanf(ss,"%s %s",c[num],s);
Build_Tree();
num++;
}
char s1[100];
while(~scanf("%s",s1))
{
int flag=Find(s1);
//cout<<flag;
if(flag)
printf("%s\n",c[flag]);
else
printf("eh\n");
}
}