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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.
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.
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
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"); } }