多校2.1012La Vie en rose

最新推荐文章于 2016-08-15 20:17:57 发布

原创最新推荐文章于 2016-08-15 20:17:57 发布 · 439 阅读

0 ·

CC 4.0 BY-SA版权

暴力专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个模式匹配问题，其中教授张希望通过给定的模式串生成尽可能多的变化模式，并在目标字符串中查找所有这些变化模式的出现位置。文章提供了一种解决方案，通过允许选择某些不相邻的字符并与其相邻字符进行交换来生成变化模式。

点击打开链接

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

(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

题意：给一个原串给一个匹配串，匹配串可以任选一些不相邻的字符，把这些字符和相邻的字符交换，匹配成功串首为1，else0；

解：标程用的bitset,不晓得是什么阿..暴力如下：

<pre name="code" class="cpp">

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;


char ans[100000 + 5];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        string a, b;
        cin>>a>>b;
        memset(ans, '1', sizeof ans);
        for(int i = 0; i <= n-m; i++)
        {

            for(int j = 0; j < m; j++)
            {
                if(a[i+j] == b[j])
                    ;
                else if(j < m-1 && a[i+j] == b[j+1] && a[i+j+1] == b[j])      // 交换匹配，先控制范围
                    j++;
                else
                {
                    ans[i] = '0';
                    break;
                }
            }

        }
        for(int i = 0; i < n-m+1; i++)
            cout<<ans[i];
        while(--m)
            cout<<'0';                                 //倒数几个长度不够的一定匹配不到。

        cout<<endl;
    }
    return 0;

}