La Vie en rose

題目連接：

http://acm.hdu.edu.cn/showproblem.php?pid=5745

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:

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

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

Hint

題意

給你一個串s，然后給你個串s2

s2可以在其中選擇出一個子序列，子序列中相鄰的元素距離至少差2，然后交換前后位置（建議讀題，我說的不是很清楚……）

再去匹配，問你有哪些位置可以匹配。

題解：

正解是dp+bitset優化

現場做的時候，看大多數隊伍都過了，就感覺應該是傻逼暴力題了，剪剪枝就過了。。。

建議還是去寫正解，正解其實也很好寫的……

代碼

#include<bits/stdc++.h> using namespace std;int n,m; char s1[100004],s2[5005],s3[5005]; int sum[100004][26]; int tmp[26]; void solve(){string ans="";memset(sum,0,sizeof(sum));memset(tmp,0,sizeof(tmp));memset(s3,0,sizeof(s3));scanf("%d%d",&n,&m);scanf("%s%s",s1+1,s2+1);s2[m+1]='.';for(int i=1;i<=n;i++){for(int j=0;j<26;j++){sum[i][j]=sum[i-1][j];}sum[i][s1[i]-'a']++;}for(int i=1;i<=m;i++)tmp[s2[i]-'a']++;for(int i=1;i<=m+1;i++)s3[i]=s2[i];for(int i=1;i<=n;i++){int flag = 1;if(i+m-1>n){ans+='0';continue;}for(int j=0;j<26;j++){if(sum[i+m-1][j]-sum[i-1][j]!=tmp[j]){flag = 0;break;}}if(flag==0){ans+='0';continue;}int ff = 0;int len = 0;for(int j=1;j<=m;j++){len = max(len,j);if(s2[j]==s1[i+j-1]){ff=0;}else if(ff==0&&s2[j+1]==s1[i+j-1]){swap(s2[j+1],s2[j]);ff=1;}else{flag=0;break;}}for(int j=1;j<=len+1;j++)s2[j]=s3[j];if(flag==0){ans+='0';continue;}ans+='1';}cout<<ans<<endl; } int main(){int t;scanf("%d",&t);while(t--)solve();return 0; }

轉載于:https://www.cnblogs.com/qscqesze/p/5692872.html

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