String distance (zifu)

Description

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For two different strings, we have a set of operation methods to make them the same,the specific method is:
Modify a character (such as replacing "\(\text{a}\)" with "\(\text{b}\)")
Delete a character (e.g. change "\(\text{traveling}\)" to "\(\text{traveing}\)")
For example, for the two strings "\(\text{abcdefg}\)" and "\(\text{abcdef}\)", we think we can achieve the goal by adding or subtracting a "\(\text{g}\)". Whether we increase or decrease "\(\text{g}\)", we only need one operation. We define the number of times required for this operation as the distance between two strings.
Given any two strings, write an algorithm to calculate their distance.

Format

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Input

Line 1 has an integer \(n\). Indicates the number of groups of test data. \((n \leq 50)\)
Next, there are a total of n lines, with two strings in each line, separated by spaces, indicating the two strings to be calculated.
The length of the string does not exceed 1000.

Output

Output an integer for each set of test data, which is the distance between two strings

Sample

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Input

3
abcdefg   abcdef
ab ab
mnklj jlknm

Output

1
0
4

Sample Code

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#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;

int t,m,n;
char a[1005],b[1005];
int f[1005][1005];//The minimum distance between two strings.

int main() {
	freopen("zifu.in","r",stdin);
	freopen("zifu.out","w",stdout);
	cin>>t;
	while(t--) {
		// scanf("%s%s",a,b);
		cin>>a>>b;   //When cin encounters a space, the input is deemed to be over. Same for scanf.
		m=strlen(a);
		n=strlen(b);
		for(int i=1; i<=m; i++) f[i][0]=i;
		for(int i=1; i<=n; i++) f[0][i]=i; //Initialization of the boundary.
		for(int i=1; i<=m; i++)
			for(int j=1; j<=n; j++)
				if(a[i-1]==b[j-1]) f[i][j]=f[i-1][j-1];//Determine whether you need to change.
				else f[i][j]=min(min(f[i-1][j],f[i][j-1]),f[i-1][j-1])+1;//Delete a[i] and insert b[j-1] after a[i]. This a[i] is b[j] and choose the one with the least number of changes.
		cout<<f[m][n]<<endl;
	}
	return 0;
}