HDU-5972 Regular Number（Shift-And）

Regular Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 880    Accepted Submission(s): 212

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 isai(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

43 0 9 72 5 72 2 52 4 509755420524

 

Sample Output

975575540524

题意：一串数字由n位组成，第i位可以有a[i]种数字。现在给出一个长字符串，问这个字符串有多少串数字
题解：Shift-And算法模板题

#include<bits/stdc++.h>using namespace std;const int MXP = 1e3 + 5;const int MXT = 5e6 + 5;char P[MXP],T[MXT];bitset<MXP>B[256],D;  //B数组大小为字符的种数,长度为MXPvoid pre_solve(char P[]){    int len=strlen(P);    D.reset();    for(int i=0;i<256;i++) B[i].reset();    for(int i=0;i<len;i++) B[P[i]].set(i);}int n;int shift_and(char T[]) {    int m=strlen(T);    //pre_solve(P);    for (int i = 0; i < m; i++) {        D=(D<<1).set(0)&B[T[i]];        if (D[n-1]) {            char ch=T[i+1];            T[i+1]=0;            puts(T+i-n+1);            T[i+1]=ch;        }    }    return -1;}char tmp[MXT];int main(){    //freopen("in.txt","r",stdin);    scanf("%d",&n);    int num, x;    for (int i = 0; i < n; i++) {        scanf("%d",&num);        for (int j = 1; j <= num; j++) {            scanf("%d",&x);            B[x+'0'].set(i);        }    }    getchar();    gets(tmp);    shift_and(tmp);    return 0;}