【HDU】 2846 Repository

Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3442    Accepted Submission(s): 1291

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

20adaeafagahaiajakaladsaddadeadfadgadhadiadjadkadlaes5badads

 

Sample Output

02011112

 

题解：直接暴力出每个字符串的子串加入Trie，最后查找。

注意abab在插入的时候ab只需要加一次，要做好记号。

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;struct Trie{    Trie *ch[27];    int val,f;};Trie *root;char c[25],s[25];int P,Q;void buildtrie(char *s,int flag){    Trie *p=root,*q;    int l=strlen(s),k;    for (int i=0;i<l;i++)    {        k=s[i]-97;        if (p->ch[k]==NULL)        {            q=(Trie*)malloc(sizeof(Trie));            for (int j=0;j<26;j++) q->ch[j]=NULL;            q->val=1; q->f=flag;            p->ch[k]=q;            p=p->ch[k];        }        else        {            if (p->f!=flag)            {                p->f=flag;                p->val++;            }            p=p->ch[k];        }    }    if (p->f!=flag)    {        p->f=flag;        p->val++;    }}void solve(char *s,int flag){    int p,l=strlen(s);    for (int i=0;i<l;i++)        for (int j=i;j<l;j++)    {        p=0;        for (int k=i;k<=j;k++,p++) c[p]=s[k];        buildtrie(c,flag);        memset(c,0,sizeof(c));    }}int intrie(char *s){    Trie *p=root;    int k,l=strlen(s);    for (int i=0;i<l;i++)    {        k=s[i]-97;        if (p->ch[k]!=NULL) p=p->ch[k];        else return 0;    }    return p->val;}int main(){    root=(Trie*)malloc(sizeof(Trie));    for (int i=0;i<26;i++) root->ch[i]=NULL;    root->val=0; root->f=0;    scanf("%d",&P);    for (int i=1;i<=P;i++)    {        scanf("%s",s);        solve(s,i);    }    scanf("%d",&Q);    while (Q--)    {        scanf("%s",s);        printf("%d\n",intrie(s));    }    return 0;}