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.
InputIt 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)ai(1≤ai≤10),representing that the i-th position of regular expression has aiai numbers to be selected.Next there are aiainumeric characters. In the last line,there is a numeric string.The length of the string is not more than 5 * 10^6.OutputOutput all substrings that can be matched by the regular expression. Each substring occupies one lineSample Input
4 3 0 9 7 2 5 7 2 2 5 2 4 5 09755420524
Sample Output
9755 7554 0524题意：有n组数，同组数之间的关系是 | ，然后输入一行字符串，问字符串有哪些子串能够用这n组数来表示？？
Input：
先输入n，然后输入n组数，每组数的第一个（ai）表示这组数的长度，然后输入ai个数，他们之间的关系是 | ，然后输入一行字符串。
Output：
输出满足条件的子串。
思路：开一个规格为10的bitset数组和一个bitset类型的test，然后根据输入存放bitset数组。输入字符串后对字符串进行遍历当test[n]==1时输出字符串[i-n+1,i]。输出是可以把下一个变成‘\0’输出字符串s+i-n+1，然后再变回来。
AC代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <algorithm>
6 #include <map>
7 #include <set>
8 #include <stdlib.h>
9 #include <stack>
10 #include <vector>
11 #include <cmath>
12 #include <bitset>
13 #define ll long long
14 using namespace std;
15 const int maxn=1005;
16 bitset<maxn>vec[10];
17 bitset<maxn>dp;
18 char s[5000005];
19 int main()
20 {
21 int n,m,x;
22 while(~scanf("%d",&n))
23 {
24 for(int i=0;i<=9;i++)
25 vec[i].reset();
26 for(int i=0;i<n;i++)
27 {
28 scanf("%d",&m);
29 for(int j=0;j<m;j++)
30 {
31 scanf("%d",&x);
32 vec[x].set(i);
33 }
34 }
35 scanf("%s",s);
36 int len=strlen(s);
37 dp.reset();
38 for(int i=0;i<len;i++)
39 {
40 dp=dp<<1;
41 dp=dp.set(0);
42 dp=dp&vec[s[i]-‘0‘];
43 if(dp[n-1]==1)
44 {
45 char c=s[i+1];
46 s[i+1]=‘\0‘;
47 printf("%s\n",s+i-n+1);
48 s[i+1]=c;
49 }
50 }
51 }
52 return 0;
53 }