Problem Description
Using regular expression to define a numeric string is a very common thing. Generally, use the shape as follows:
(0|9|7) (5|6) (2) (4|5)
Above regular expression matches 4 digits:The first is one of 0,9 and 7. The second is one of 5 and 6. The third is 2. And the fourth is one of 4 and 5. The above regular expression can be successfully matched to 0525, but it cannot be matched to 9634.
Now,giving you a regular expression like the above formula,and a long string of numbers,please find out all the substrings of this long string that can be matched to the regular expression.
Input
It contains a set of test data.The first line is a positive integer N (1 ≤ N ≤ 1000),on behalf of the regular representation of the N bit string.In the next N lines,the first integer of the i-th line is ai(1≤ai≤10),representing that the i-th position of regular expression has ai numbers to be selected.Next there are ai numeric characters. In the last line,there is a numeric string.The length of the string is not more than 5 * 10^6.
Output
Output all substrings that can be matched by the regular expression. Each substring occupies one line
Sample Input
4
3 0 9 7
2 5 7
2 2 5
2 4 5
09755420524
Sample Output
9755
7554
0524
Source
2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）
题意：输入N ，表示有一个长为N的子串，接下来N行，每行输入ai 和ai个数，表示有ai个数，表示子串第i个字符为ai个数中的一个，也就是这个子串的正则式，然后输入一个主串，要求输出这个主串中包含的所有的子串。
思路：使用bitset<N> b[10] ，b[i][j]表示值为i的数可以出现在子串的那些位置，即下标，那么对主串进行遍历 i=0:len-1 。另外定义一个变量bitset<N> ans ,每次先将ans左移一位，然后将最低位置1，然后令k=当前输入的数，将ans=ans&b[k], 如果当前ans[N-1]==1,那么主串s[i-N+1]~s[i]就是合法子串，输出；
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <bitset>
using namespace std;
typedef long long LL;
const int M=5*1e6+5;
bitset<1005>b[12];
bitset<1005>ans;
char s[5000005];
int main()
{
int N;
while(scanf("%d",&N)!=EOF)
{
for(int i=0;i<10;i++) b[i].reset();
for(int i=0;i<N;i++)
{
int n;
scanf("%d",&n);
for(int j=1;j<=n;j++)
{
int k;
scanf("%d",&k);
b[k].set(i);
}
}
getchar();
gets(s);
ans.reset();
int len=strlen(s);
for(int i=0;i<len;i++)
{
ans=ans<<1;
ans.set(0);
ans=ans&b[s[i]-‘0‘];
if(ans[N-1]==1){
char c=s[i+1];
s[i+1]=‘\0‘;
puts(s+i-N+1);
s[i+1]=c;
}
}
}
return 0;
}