hdu 5745 La Vie en rose（2016多校第二场）

La Vie en rose

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 643    Accepted Submission(s):
328

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

题意：输入两个字符串s串和p串，p串从s串的第一个字符开始一直比较到最后一个（如果后面字符比p串段，则肯定输出0），比较的时候若不相同，则p串的字符可以相邻交换进行比较，每个位置只能交换一次，（只是当前比较的交换，而不是永久交换），相同字符串输出1，不同则输出0。

纯暴力，从第一个字符开始比较就好。

附上代码：

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #define  ll long long
 using namespace std;
 char s1[],s2[];
 int main()
 {
     int T,i,j;
     scanf("%d",&T);
     while(T--)
     {
         int len1,len2;
         memset(s1,'\0',sizeof(s1));
         memset(s2,'\0',sizeof(s2));
         scanf("%d%d",&len1,&len2);
         scanf("%s%s",s1,s2);
         char w;
         if(len1<len2)
         {
             for(i=; i<len1; i++)
                 printf("");
             printf("\n");
             continue;
         }
         for(i=; i<len1; i++)
         {
             int t=;
             for(j=i; j<i+len2&&j<len1; j++)
             {
                 if(s1[j]!=s2[t])
                 {
                     if(j==i+len2-)
                         break;
                         if(s1[j+]!=s2[t]||s1[j]!=s2[t+])
                             break;
                         else
                         {
                             t+=;
                             j+=;
                         }
                 }
                 else
                 {
                     t++;
                 }
             }
             if(j==i+len2)
                 printf("");
             else
                 printf("");

         }
         printf("\n");
     }
     return ;
 }