CF1705D Mark and Lightbulbs

Mark and Lightbulbs - 洛谷

Mark has just purchased a rack of nn lightbulbs. The state of the lightbulbs can be described with binary string s = s_1s_2\dots s_ns=s1​s2​…sn​ , where s_i=\texttt{1}si​=1 means that the ii -th lightbulb is turned on, while s_i=\texttt{0}si​=0 means that the ii -th lightbulb is turned off.

Unfortunately, the lightbulbs are broken, and the only operation he can perform to change the state of the lightbulbs is the following:

• Select an index ii from 2,3,\dots,n-12,3,…,n−1 such that s_{i-1}\ne s_{i+1}si−1​=si+1​ .
• Toggle s_isi​ . Namely, if s_isi​ is \texttt{0}0 , set s_isi​ to \texttt{1}1 or vice versa.

Mark wants the state of the lightbulbs to be another binary string tt . Help Mark determine the minimum number of operations to do so.

输入格式

The first line of the input contains a single integer qq ( 1\leq q\leq 10^41≤q≤104 ) — the number of test cases.

The first line of each test case contains a single integer nn ( 3\leq n\leq 2\cdot 10^53≤n≤2⋅105 ) — the number of lightbulbs.

The second line of each test case contains a binary string ss of length nn — the initial state of the lightbulbs.

The third line of each test case contains a binary string tt of length nn — the final state of the lightbulbs.

It is guaranteed that the sum of nn across all test cases does not exceed 2\cdot 10^52⋅105 .

输出格式

For each test case, print a line containing the minimum number of operations Mark needs to perform to transform ss to tt . If there is no such sequence of operations, print -1−1 .

输入输出样例

输入 #1复制

4
4
0100
0010
4
1010
0100
5
01001
00011
6
000101
010011

输出 #1复制

2
-1
-1
5

说明/提示

In the first test case, one sequence of operations that achieves the minimum number of operations is the following.

• Select i=3i=3 , changing \texttt{01}\color{red}{\texttt{0}}\texttt{0}0100 to \texttt{01}\color{red}{\texttt{1}}\texttt{0}0110 .
• Select i=2i=2 , changing \texttt{0}\color{red}{\texttt{1}}\texttt{10}0110 to \texttt{0}\color{red}{\texttt{0}}\texttt{10}0010 .

In the second test case, there is no sequence of operations because one cannot change the first digit or the last digit of ss .In the third test case, even though the first digits of ss and tt are the same and the last digits of ss and tt are the same, it can be shown that there is no sequence of operations that satisfies the condition.

In the fourth test case, one sequence that achieves the minimum number of operations is the following:

• Select i=3i=3 , changing \texttt{00}\color{red}{\texttt{0}}\texttt{101}000101 to \texttt{00}\color{red}{\texttt{1}}\texttt{101}001101 .
• Select i=2i=2 , changing \texttt{0}\color{red}{\texttt{0}}\texttt{1101}001101 to \texttt{0}\color{red}{\texttt{1}}\texttt{1101}011101 .
• Select i=4i=4 , changing \texttt{011}\color{red}{\texttt{1}}\texttt{01}011101 to \texttt{011}\color{red}{\texttt{0}}\texttt{01}011001 .
• Select i=5i=5 , changing \texttt{0110}\color{red}{\texttt{0}}\texttt{1}011001 to \texttt{0110}\color{red}{\texttt{1}}\texttt{1}011011 .
• Select i=3i=3 , changing \texttt{01}\color{red}{\texttt{1}}\texttt{011}011011 to \texttt{01}\color{red}{\texttt{0}}\texttt{011}010011 .

//非常精巧的一个题目，发现的规律是，一个置换无论进行多少次，都不会消灭一段0或者1，即s有多少段,t就有多少段。而且我们会发现，置换一次会将01分界线偏移一位，在排除无解（首位不同）的情况下，只需要求出每一个01分界点的坐标即可，在满足首位相同，01分界点个数相同的情况下，我们s,t的01段顺序就已经确保完全一致了，所以没有必要再去判断是否01段顺序不同

# include
# include
 
using namespace std;
typedef long long int ll;
int a[200000+10],b[200000+10];
 
int main ()
{
     int t;
 
     cin>>t;
 
     while(t--)
     {
         int n;
 
         cin>>n;
 
         string s,t;
 
         cin>>s>>t;
 
         if(s[0]!=t[0]||s[s.length()-1]!=t[t.length()-1])
         {
             cout<<-1<<'\n';
             continue;
 
         }
         int len1=0,len2=0;
 
         for(int i=1;i
         {
             if(s[i]!=s[i-1])
             {
                 len1++;
 
                 a[len1]=i;
 
             }
 
 
             if(t[i]!=t[i-1])
             {
                 len2++;
 
                 b[len2]=i;
 
             }
 
         }
 
         if(len1!=len2)
         {
             cout<<-1<<'\n';
             continue;
 
         }
 
         ll ans=0;
         for(int i=1;i<=len1;i++)
         {
             ans+=abs(a[i]-b[i]);
         }
 
         cout<
 
     }
    return 0;
 
 
}