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最新推荐文章于 2019-01-22 11:31:45 发布

蒟蒻的ACMer

最新推荐文章于 2019-01-22 11:31:45 发布

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本文链接：https://blog.youkuaiyun.com/zp___waj/article/details/49532343

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本文介绍了一种使用前缀树（Trie树）进行字符串匹配的方法，通过构建商品名称的前缀树并存储其子串，实现高效搜索含有特定子串的商品。文章详细解释了Trie树的创建和查询过程，并提供了完整的C++实现代码。

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Repository

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3351 Accepted Submission(s): 1256

Problem Description

When you go shopping, you can search in repository for avalible merchandises by the computers and internet. First you give the search system a name about something, then the system responds with the results. Now you are given a lot merchandise names in repository and some queries, and required to simulate the process.

Input

There is only one case. First there is an integer P (1<=P<=10000)representing the number of the merchanidse names in the repository. The next P lines each contain a string (it's length isn't beyond 20,and all the letters are lowercase).Then there is an integer Q(1<=Q<=100000) representing the number of the queries. The next Q lines each contains a string(the same limitation as foregoing descriptions) as the searching condition.

Output

For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.

Sample Input

  
   20
ad
ae
af
ag
ah
ai
aj
ak
al
ads
add
ade
adf
adg
adh
adi
adj
adk
adl
aes
5
b
a
d
ad
s

Sample Output

Source

2009 Multi-University Training Contest 4 - Host by HDU

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gaojie | We have carefully selected several similar problems for you: 2852 2847 2845 2850 2851

给出n个串然后再询问q个串在前n个串出现的次数

因为只要出现就算那么建树的时候把串的子串也存入

因为会出现一个串中有重复出现查询串的情况所以建树的时候要标记一下

G++会超内存c++不会但是会稍微慢一点

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 100005
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
#define MAX 26
using namespace std;
struct Trie{
    Trie *next[MAX];
    int cnt;
    int flag;
};
Trie *root;
int cnt,n,q;
void creatTrie(char *str,int st,int len,int loop){
    Trie *p=root,*q;
    FOR(i,st,len-1){
        int id=str[i]-'a';
        if(p->next[id]==NULL){
            q=(Trie*)malloc(sizeof(Trie));
            q->cnt=1;
            q->flag=loop;
            FOR(j,0,MAX-1)q->next[j]=NULL;
            p->next[id]=q;
            p=p->next[id];
        }
        else {
                p=p->next[id];
                if(p->flag!=loop){
                        p->flag=loop;
                p->cnt++;
                }
        }
    }
}
int  find_Trie(char *str){
    int len=strlen(str);
    Trie *p=root;
    FOR(i,0,len-1){
        int id=str[i]-'a';
        p=p->next[id];
        if(p==NULL)return 0;
    }
    return p->cnt;
}
void deal(Trie *t){
    if(t==NULL)return ;
    FOR(i,0,MAX-1)
        if(t->next[i])
            deal(t->next[i]);
    free(t);
}
char str[50];
int main(){
    while(rd(n)!=EOF){
        root=(Trie *)malloc(sizeof(Trie));
        FOR(i,0,MAX-1)root->next[i]=NULL;
        FOR(i,0,n-1){
            rds(str);
            int len=strlen(str);
            for(int j=0;j<len;++j)
            creatTrie(str,j,len,i);
            }
        rd(q);FOR(i,1,q){rds(str);cout<<find_Trie(str)<<'\12';}
        deal(root);
    }
    return 0;
}
/*
20
ad
ae
af
ag
ah
ai
aj
ak
al
ads
add
ade
adf
adg
adh
adi
adj
adk
adl
aes
5
b
a
d
ad
s
 */