1371 - Period

Given two strings A and B over an alphabet ∑ , the edit distance between A and B is the minimum number of edit operations needed to convert A into B . The three edit operations are the following:

(i)

change: replace one character of A by another single character of B .

(ii)

deletion: delete one character from A .

(iii)

insertion: insert one character of B into A .

For example, the following figure shows that the edit distance between the strings A =abcdefg and B =ahcefig is 3. The edit operations are a change (i.e., replacing b of A by h of B ), a deletion (i.e., deleting d from A ), and an insertion (i.e., inserting i of B into A ).

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We now define a period of a repetitive string as follows: The string p is called the exact period of a string x if x can be written as x = pk , where k≥1 and p is the shortest string. For example, if x =abababab then x = (abababab)1 = (abab)2 = (ab)4 . Thus, the string ab is the exact period of x . We define an approximate period similarly. Given two strings x and y , suppose that the string x is partitioned into substrings pi , 1≤i≤t , where pi is not a null string, i.e., x = p1 . p2 . p3 … pt . If the edit distance between a string y and each substring pi is less than or equal to an integer k , string y is called a k -approximate period of string x . In this problem, given two strings x and y , we want to find the minimum k such that string y is a k -approximate period of string x . For example, suppose that two strings x = abcdabcabb and y =abc are given. Since x may be partitioned into x = p1 . p2 . p3 = abcd . abc . abb and the edit distances between string y =abc and each substring abcd, abc, and abb equal to 1, 0, and 1, respectively, y is a 1-approximate period of x . Hence, the minimum k is one.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. For each test case, a string y is given in the first line and the string x is given in the next line. The length of string y is at least 1 and at most 50, the length of string x is at least 1 and at most 5000, and the alphabet ∑ is the set of lowercase English characters.

Output

Your program is to write to standard output. Print exactly one line for each test case. Print the minimum integer value k such that string y is a k -approximate period of string x . The following shows sample input and output for three test cases.

Sample Input

3

abc

abcdabcabb

abab

abababababab

xyz

abcdefghikjlmn

Sample Output

1

0

3

我有话说：

大致思路是二分加DP。

二分答案，动态规划判断。这个问题可以参考LCS问题。

不过LCS问题求的是相同的。该题应当在判断a[i+1]和b[j+1]是否不同。不同就必须操作一次（修改）。+1。还有插入操作。

最重要的是if(dp[i][m]<=key)dp[i][0]=0;这条语句。就是在a[i]这断开。重新比较。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=5000+10;
const int maxm=50+10;
const int INF=0x3f3f3f3f;
int n,m,dp[maxn][maxm];
char a[maxn],b[maxm];
bool judge(int key){
memset(dp,INF,sizeof(dp));
dp[0][0]=0;
for(int i=0;i<=n;i++){
if(dp[i][m]<=key)dp[i][0]=0;//划分
for(int j=0;j<=m;j++){
if(dp[i][j]>key)continue;
dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j]+(a[i+1]==b[j+1]?0:1));//修改
dp[i][j+1]=min(dp[i][j+1],dp[i][j]+1);//插入
dp[i+1][j]=min(dp[i+1][j],dp[i][j]+1);
}
}
return dp
[m]<=key;
}
int solve(){
int L,R;
L=0,R=m;
while(L<R){
int mid=(L+R)/2;
if(judge(mid))R=mid;
else L=mid+1;
}
return L;
}
int main()
{
int T;
cin>>T;
while(T--){
scanf("%s%s",b+1,a+1);
n=strlen(a+1);
m=strlen(b+1);
printf("%d\n",solve());
}
return 0;
}