Codeforces Round #599 (Div. 2) B2. Character Swap (Hard Version) 构造

Link: https://www.luogu.com.cn/problem/CF1243B2 

Topic: To give you two strings s and t of length n, you can perform up to 2 * n operations, select i and j for each operation, and then exchange s [i] and t [j], ask if you can Whether to make the two strings the same

Construction method: If the (0 ~ i) parts s and t are already equal, at the i position, first find whether there is the same character as t [i] in (i + 1 ~ t.size ()-1), if found , Then exchange s [i] and t [j], if not found, then find in s, after finding, first exchange s [j] and t [t.size ()-1], then exchange s [i] and t [t.size ()-1] exchange this number of times at most 2 * n times; then s = t;

Code:

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=3e5+10;
int n;
string s,t;
vector<pair<int ,int> > v;
void solve()
{
    cin>>n>>s>>t;
    v.clear();
    for(int i=0; i<s.size(); i++)
    {
     if(s[i]!=t[i])
        {
            int flag=0;
            for(int j=i+1; j<t.size(); j++)
            {
                if(t[i]==t[j])
                    {
                        flag=1;
                        v.push_back(make_pair(i+1,j+1));
                        swap(s[i],t[j]);
                        break;
                    }
            }
        if(!flag)
        {
            for(int j=i+1; j<s.size(); j++)
            {
                if(t[i]==s[j])
                {
                    flag=1;
                    v.push_back(make_pair(j+1,t.size()));
                    swap(s[j],t[t.size()-1]);
                    v.push_back(make_pair(i+1,t.size()));
                    swap(s[i],t[t.size()-1]);
                    break;
                }
            }
        }
        if(!flag)
        {
            puts("NO\n");
            return ;
        }
    }
    }
    puts("YES\n");
    cout<<v.size()<<endl;
    for(int i=0; i<v.size(); i++)
        cout<<v[i].first<<" "<<v[i].second<<endl;
    return ;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}