Repository
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 88 Accepted Submission(s) : 18
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
0
20
11
11
2
Source
2009 Multi-University Training Contest 4 - Host by HDU
此题和之前的那道“统计难题”很像,所以这里我们仍然采用数组模拟的方法来解决这道题,如果之前“统计难题”看懂了做出来了,那这道题对于我们来说,就是略微加强一点的水题啦(小编~)。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
using namespace std;
struct node
{
int id;
int cnt;
int al[26];
};
int ptr=0;
node word[1000000];
void Init(int z)
{
for(int i=0;i<26;i++)
word[z].al[i]=-1;
}
void Insert(char*s,int len,int x)
{
int now=0;
for(int i=0;i<len;i++)
{
if(word[now].al[s[i]-'a']==-1)
{
word[now].al[s[i]-'a']=++ptr;
Init(ptr);
now=ptr;
}
else now=word[now].al[s[i]-'a'];
if(word[now].id!=x) word[now].cnt++;
word[now].id=x;
}
}
int find(char*s)
{
int now=0;
int len=strlen(s);
for(int i=0;i<len;i++)
{
if(word[now].al[s[i]-'a']==-1)
return 0;
now=word[now].al[s[i]-'a'];
}
return word[now].cnt;
}
int main()
{
int n,m,len;
char s[21];
Init(0);
scanf("%d",&n);
while(n--)
{
scanf("%s",&s);
len=strlen(s);
for(int i=0;i<len;i++)
Insert(s+i,len-i,n+1);
}
scanf("%d",&m);
while(m--)
{
scanf("%s",&s);
printf("%d\n",find(s));
}
return 0;
}