Little jay really hates to deal with string. But moondy likes it very much, and she's so mischievous that she often gives jay some dull problems related to string. And one day, moondy gave jay another problem, poor jay finally broke out and cried, " Who can help me? I'll bg him! "
So what is the problem this time?
First, moondy gave jay a very long string A. Then she gave him a sequence of very short substrings, and asked him to find how many times each substring appeared in string A. What's more, she would denote whether or not founded appearances of this substring are allowed to overlap.
At first, jay just read string A from begin to end to search all appearances of each given substring. But he soon felt exhausted and couldn't go on any more, so he gave up and broke out this time.
I know you're a good guy and will help with jay even without bg, won't you?
Input
Input consists of multiple cases( <= 20 ) and terminates with end of file.
For each case, the first line contains string A ( length <= 10^5 ). The second line contains an integer N ( N <= 10^5 ), which denotes the number of queries. The next N lines, each with an integer type and a string a ( length <= 6 ), type = 0 denotes substring a is allowed to overlap and type = 1 denotes not. Note that all input characters are lowercase.
There is a blank line between two consecutive cases.
Output
For each case, output the case number first ( based on 1 , see Samples ).
Then for each query, output an integer in a single line denoting the maximum times you can find the substring under certain rules.
Output an empty line after each case.
Sample Input
ab
2
0 ab
1 ab
abababac
2
0 aba
1 aba
abcdefghijklmnopqrstuvwxyz
3
0 abc
1 def
1 jmn
Sample Output
Case 1
1
1
Case 2
3
2
Case 3
1
1
0
/*
ZOJ 3228 Searching the String(AC自动机)
给你几个子串,然后在字符串中查询它们出现的次数.但是0表示可以重复,1表示不可以
重复.
在开始想的是建两个然后分别查询.但是发现完全可以一次查询解决 TAT
abababac
2
0 aba
1 aba
就这一组数据而言.
建成：
root
/
①a
/
②b
/
③a
对于可以重复的部分,直接进行查找就行. 因为叶子节点的a的nex[a][b]就是它的父亲
b节点
//可以参考’飘过的小牛‘的总结,主要是fail指针的理解
所以导致 a① -> b② -> a③ -> b② -> a③ 时又走到了叶子节点a.
而且只有到走到一个字符串的终点的时候才可能 +1
于是乎在每次走完一个子串的时候(通过ed判断) 判断一下它最近一次出现的位置
两个的差是否大于子串的长度即可
hhh-2016-04-26 20:19:35
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <functional>
#include <algorithm>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef unsigned long long ll;
typedef unsigned int ul;
const int mod = 20090717;
const int INF = 0x3f3f3f3f;
const int N = 100005*6;
int pos[100005];
char str[100005];
struct Matrix
{
int len;
int ma[111][111];
Matrix() {};
Matrix(int L)
{
len = L;
}
};
struct Tire
{
int nex[N][26],fail[N],ed[N];
int dep[N];
int tan[N][2];
int root,L;
int newnode()
{
for(int i = 0; i < 26; i++)
nex[L][i] = -1;
ed[L++] = 0;
return L-1;
}
void ini()
{
L = 0;
root = newnode();
dep[root] = 0;
}
int cal(char ch)
{
if(ch == 'A')
return 0;
else if(ch == 'C')
return 1;
else if(ch == 'G')
return 2;
else if(ch == 'T')
return 3;
}
int inser(char buf[])
{
int len = strlen(buf);
int now = root;
for(int i = 0; i < len; i++)
{
int ta = buf[i] - 'a';
if(nex[now][ta] == -1)
{
nex[now][ta] = newnode();
dep[nex[now][ta]] = i+1;
}
now = nex[now][ta];
}
ed[now] ++;
return now;
}
void build()
{
queue<int >q;
fail[root] = root;
for(int i = 0; i < 26; i++)
if(nex[root][i] == -1)
nex[root][i] = root;
else
{
fail[nex[root][i]] = root;
q.push(nex[root][i]);
}
while(!q.empty())
{
int now = q.front();
q.pop();
// if(ed[fail[now]])
// ed[now] = ed[fail[now]];
for(int i = 0; i < 26; i++)
{
if(nex[now][i] == -1)
nex[now][i] = nex[fail[now]][i];
else
{
fail[nex[now][i]] = nex[fail[now]][i];
q.push(nex[now][i]);
}
}
}
}
Matrix to_mat()
{
Matrix mat(L);
memset(mat.ma,0,sizeof(mat.ma));
for(int i = 0; i < L; i++)
{
for(int j = 0; j < 4; j++)
{
if(!ed[nex[i][j]])
mat.ma[i][nex[i][j]] ++;
}
}
return mat;
}
int last[N];
void query(char buf[])
{
int len = strlen(buf);
int cur = root;
memset(tan,0,sizeof(tan));
memset(last,-1,sizeof(last));
for(int i = 0;i < len;i++)
{
int ta = buf[i]-'a';
cur = nex[cur][ta];
int t = cur;
while(t != root)
{
if(ed[t])
{
tan[t][0]++;
if(i-last[t] >= dep[t])
{
last[t] = i;
tan[t][1] ++;
}
}
t = fail[t];
}
}
return ;
}
};
Tire ac;
char s[10];
int ty[100004];
int main()
{
int cas = 1;
int n;
while(scanf("%s",str) != EOF)
{
scanf("%d",&n);
ac.ini();
printf("Case %d
",cas++);
for(int i = 0; i < n; i++)
{
scanf("%d%s",&ty[i],s);
pos[i] = ac.inser(s);
}
ac.build();
ac.query(str);
// for(int i = 0;i < n;i++)
// cout << pos[i] <<" ";
// cout <<endl;
for(int i = 0;i < n;i++)
{
printf("%d
",ac.tan[pos[i]][ty[i]]);
}
printf("
");
}
return 0;
}