poj-2503-Babelfish-字典树orMAP or 字符串哈希

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题意:

给你一组字符串，每一行有两个串，a,b；

输入空行结束

给你一个串c，问你在前面的第二列的数组中出没出现过，有的话，输出前面的那个，没有的话输出eh

sscanf函数应用是，把一个串，分割成不同类型的

1. map 2625ms

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<climits>#include<list>#define MULT 20000using  namespace std;int main(){   char ls[30], lk[30],lp[30];   map<string,string>mapS;   while(gets(ls))   {       if(strlen(ls)==0)        break;       sscanf(ls,"%s %s",lk,lp);       mapS[lp] = lk;   }   map<string,string>::iterator iter;   while(gets(ls))   {       iter = mapS.find(ls);       if(iter!=mapS.end())       {            cout<<iter->second<<endl;       }       else        printf("eh\n");   }   return 0;}

2 哈希 813ms

#include<stdio.h>#include<string.h>#include<stdlib.h>#define mod 2000int tmp ;using  namespace std;struct node{  char s[30];  char a[30];  int next;}ls[100500];int head[2500];int hash_index(char s[]){    int hash=1;    int len=strlen(s);    for(int i=0; i<len; i++)        hash=(hash*29+s[i]-'a')%mod;    return hash;}int main(){    tmp = 1;    char tr[30], tl[30], to[30];    memset(head,-1,sizeof(head));    while(gets(to))    {        if(strlen(to)==0)            break;        sscanf(to,"%s %s",tl,tr);        int key = hash_index(tr);        if(head[key]==-1)        {            strcpy(ls[tmp].s,tl);            strcpy(ls[tmp].a,tr);            ls[tmp].next = -1;            head[key] = tmp++;        }        else        {            int c = head[key];            while(1)            {                if(ls[c].next!=-1)                    c = ls[c].next;                else                    break;            }            strcpy(ls[tmp].s,tl);            strcpy(ls[tmp].a,tr);            ls[tmp].next = -1;            ls[c].next = tmp++;        }    }    while(gets(tr))    {        int key = hash_index(tr);        int c = head[key];         if(head[key]==-1)        {            printf("eh\n");            continue;        }        while(c!=-1)        {            if(strcmp(ls[c].a,tr)==0)            {                printf("%s\n",ls[c].s);                break;            }           c = ls[c].next;        }        if(c == -1)            printf("eh\n");    }   return 0;}

3. 字典树 532ms

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<iostream>using namespace std;int top;struct node{    int next[30];    int n;    char s[30];} pos[150500];char ls[35];void add(char st[],int flag){    int j = 0, len = strlen(st);    int i;    for( i = 0; i < len; i++)    {        int x = st[i]-'a'+1;        if(pos[j].next[x]==0)            pos[j].next[x]=++top;        j = pos[j].next[x];    }    if(flag == 1)    {    if(pos[j].n == 0)    {        pos[j].n = 1;        strcpy(pos[j].s,ls);    }    }    else    {        if(pos[j].n== 0)            printf("eh\n");        else            printf("%s\n",pos[j].s);    }}int main(){    top = 0;    memset(pos,0,sizeof(pos));    char str[35], lk[35];    while(gets(str))    {        if(strlen(str)==0)            break;        sscanf(str,"%s %s",ls,lk);        add(lk,1);    }    while(gets(str))    {       add(str,0);    }    return 0;}

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