当前位置:首页 > 开发语言/框架 > C++

hdu 5972-Regular Number(字符串匹配)

优良自学吧提供hdu 5972-Regular Number(字符串匹配),hdu 5972---Regular Number(字符串匹配)题目链接   Problem Description Using regular expression to define a numeric st

hdu 5972---Regular Number(字符串匹配)

题目链接

 

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(1ai10),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;
}

 


(本文来自互联网,不代表搜站(http://www.ylzx8.cn/)的观点和立场)
本站所有内容来自互联网,若本站收录的信息无意侵犯了贵司版权,请给我们来信(ylzx8cn@163.com),我们会及时处理和回复,谢谢