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BZOJ 1819: [JSOI]Word Query电子字典

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1819

分析：
暴力枚举需要添加、删除以及替换的字符，然后建立trie树进行判断是否存在该字符串，然后利用trie树存的信息进行判重

时间貌似有点慢。

329611

yejinru

1819

Accepted

21468 kb

1232 ms

C++/Edit

2193 B

2012-12-16 17:34:39

#include <cstdio>
#include <iostream>
#include <cstring>
 
using namespace std;
 
const int kind = 26;
const int X = 25;
 
#define debug puts("here");
 
int m,n,tot;
 
struct trie{
    trie *p[kind];
    int is,id; //is 用于判重，id用于标记是否有单词在这个节点结束
    trie(){
        memset(p,NULL,sizeof(p));
        is = id = 0;
    }
};
 
void insert(trie *t,char *s){
    int id;
    for(int i=0;s[i];i++){
        id = s[i]-'a';
        if(t->p[id]==NULL)
            t->p[id] = new trie();
        t = t->p[id];
    }
    t->is = t->id = tot;
}
 
int has(trie *t,char *s){ //判断是否存在字符串s
    int c = 0;
    for(int i=0;s[i];i++){
        c = s[i]-'a';
        if(t->p[c]==NULL)
            return 0;
        t = t->p[c];
    }
    if(t->id)
        return 1;
    return 0;
}
 
int find(trie *t,char *s,int id){ //用于判断是否存在字符串s并且s没被统计过，id在这用于判重
    int c = 0;
    for(int i=0;s[i];i++){
        c = s[i]-'a';
        if(t->p[c]==NULL)
            return 0;
        t = t->p[c];
    }
    if(t->id==0||t->is==id) //不存在单词或者已经算过一次了
        return 0;
    t->is = id;
    return 1;
}
 
int main(){
    int n , m;
    while(cin>>n>>m){
        tot = 10000000;
 
        trie * root = new trie();
        char s[X],str[X];
         
        for(int i=0;i<n;i++){
            scanf("%s",s);
            insert(root,s);
        }
         
        while(m--){
            int ans = 0;
            scanf("%s",str);
 
            if(has(root,str)){
                puts("-1");
                continue;
            }
 
            tot --;
 
            int len = strlen(str);
 
            for(int i=0;i<len;i++){  // delete
                strcpy(s,str);
                 
                for(int j=i+1;j<len;j++)
                    s[j-1] = str[j];
                s[len-1] = '\0';
                 
                ans += find(root,s,tot);
            }
 
            for(int i=0;i<len;i++){  // change
                for(int j=0;j<26;j++){
                    if(str[i]==j+'a')
                        continue;
 
                    strcpy(s,str);
                    s[i] = j+'a';
                     
                    ans += find(root,s,tot);
                }
            }
 
            for(int i=0;i<=len;i++){ // add
                for(int j=0;j<26;j++){
                     
                    s[i] = j+'a';
                    for(int k=0;k<i;k++)
                        s[k] = str[k];
                    for(int k=i;k<len;k++)
                        s[k+1] = str[k];
                    s[len+1] = '\0';
 
                    ans += find(root,s,tot);
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}