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C. Double-ended Strings time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:

if |a|>0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a2a3…an; if |a|>0, delete the last character of the string a, that is, replace a with a1a2…an−1; if |b|>0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b2b3…bn; if |b|>0, delete the last character of the string b, that is, replace b with b1b2…bn−1. Note that after each of the operations, the string a or b may become empty.

For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:

delete the first character of the string a ⇒ a="ello" and b="icpc"; delete the first character of the string b ⇒ a="ello" and b="cpc"; delete the first character of the string b ⇒ a="ello" and b="pc"; delete the last character of the string a ⇒ a="ell" and b="pc"; delete the last character of the string b ⇒ a="ell" and b="p". For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.

Input

The first line contains a single integer t (1≤t≤100). Then t test cases follow.

The first line of each test case contains the string a (1≤|a|≤20), consisting of lowercase Latin letters.

The second line of each test case contains the string b (1≤|b|≤20), consisting of lowercase Latin letters.

Output

For each test case, output the minimum number of operations that can make the strings a and b equal.

思路：

暴力，先将a中的每一子串用map存下其对应的长度，同时判断b中的每一子串与a子串相同的最大长度，记为maxx，最后答案用a的长度+b的长度-2*maxx；

#include <bits/stdc++.h>
using namespace std;
#define ll long long
map<string,int>visa;
map<int,string>visb;
int main()
{
	int t;
	scanf("%d",&t);
	int cnt = 1;
	while(t--)
	{
//		printf("(%d)\n",cnt++);
		visa.clear();
		visb.clear();
		string a,b;
		cin>>a>>b;
		int ans = 0;
		int lena = a.size(),lenb = b.size();
		for(int i = 0; i <= lena; i++)
		{
			for(int j = 0; j <= lena; j++)
			{
				visa[a.substr(i,j+1)] = a.substr(i,j+1).size();
//				cout<<'('<<a.substr(i,j+1)<<')'<<a.substr(i,j+1).size()<<endl;
			}
		}
		for(int i = 0; i <= lenb; i++)
		{
			for(int j = 0; j <= lenb; j++)
			{
				if(visa[b.substr(i,j+1)] > ans)
				ans = max(ans,visa[b.substr(i,j+1)]);
			}
		}
		printf("%d\n",lena+lenb-2*ans);
	}
}