hdu5972_Regular Number_快速匹配+bitset

Regular Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 789    Accepted Submission(s): 181

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：

给你N位数，接下来有N行，第i行先输入n，表示这个数的第i 位上可以在接下来的n个数中挑选，然后i 行再输n个数。

然后输入需要匹配的母串，让你输出母串中有多少个可行的N位子串。

解：

这题首先没法使用KMP，因为在匹配失败后没法返回到准确的位置。

然后在网上向别人学了下代码，才明白这题bitset的巧妙的运用。

关于bitset的用法：http://blog.csdn.net/no2015214099/article/details/72902794

这题 bitset 的使用相当于是作为一个指针来使用的。

首先用bitset定义出现的数会在哪几位上出现，置为1。

定义ans的初始位为1，每一次母串对应位与该位出现的数的bitset进行与比较（表明该位上是否能出现该数）。因为一旦失败则置0，因此如果1出现在ans的第n位上则表明之前的n-1位全部匹配成功。

此题，使用bitset的复杂度为O(n*len/x)（len为母串长，x为机器码长）。

此题必须使用puts，gets进行输入输出，不然会超时。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <bitset>using namespace std;const int maxn=1e6+50;char str[5*maxn];bitset<1005> s[20];bitset<1005> ans;int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<20;i++)        s[i].reset();        ans.reset();        for(int i=0;i<n;i++)        {            int N,temp;            scanf("%d",&N);            for(int j=0;j<N;j++)            {                scanf("%d",&temp);                s[temp].set(i);            }        }        getchar();        gets(str);        int len=strlen(str);        for(int i=0;i<len;i++)        {            ans=ans<<1;            ans[0]=1;            ans&=s[str[i]-'0'];            if(ans[n-1]==1)            {                char temp=str[i+1];                str[i+1]='\0';                puts(str+i-n+1);                str[i+1]=temp;            }        }    }    return 0;}