2016 Multi-University Training Contest 2 La Vie en rose

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Problem Description

Professor Zhang would like to solve the multiple pattern matching problem, but he only has only one pattern string  p=p1p2...pm. So, he wants to generate as many as possible pattern strings from  p using the following method:

1. select some indices  i1,i2,...,ik such that  1≤i1<i2<...<ik<|p| and  |ij−ij+1|>1 for all  1≤j<k.
2. swap  pij and  pij+1 for all  1≤j≤k.

Now, for a given a string  s=s1s2...sn, Professor Zhang wants to find all occurrences of all the generated patterns in  s.

 

Input

There are multiple test cases. The first line of input contains an integer  T, indicating the number of test cases. For each test case:

The first line contains two integers  n and  m  (1≤n≤105,1≤m≤min{5000,n}) -- the length of  s and  p.

The second line contains the string  s and the third line contains the string  p. Both the strings consist of only lowercase English letters.

 

Output

For each test case, output a binary string of length  n. The  i-th character is "1" if and only if the substring  sisi+1...si+m−1 is one of the generated patterns.

 

Sample Input

3 4 1 abac a 4 2 aaaa aa 9 3 abcbacacb abc

 

Sample Output

1010 1110 100100100

 

Author

zimpha

 

Source

2016 Multi-University Training Contest 2

暴力水过

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char a[100005],b[5005];
int ans[100005],vis[5005];
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        memset(ans,0,sizeof ans);
        memset(vis,0,sizeof vis);
        scanf("%d%d",&n,&m);
        scanf("%s%s",a,b);
        for(int i=0;i<n;)
        {
            int per=i+1;
            for(int j=0;j<m;)
            {
                if(a[i]==b[j])
                {
                    i++,j++;
                }
                else if(a[i]==b[j+1]&&a[i+1]==b[j]&&j<m&&i<n)
                {
                    i+=2;
                    j+=2;
                }
                else if(a[i]!=b[j])
                {
                    i=per;
                    break;
                }
                if(j==m)
                {
                    i=per;
                    ans[per-1]=1;
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            printf("%d",ans[i]);
        }
            printf("\n");
    }
}