Lexicographically Small Enough(平衡树)

Posted 2022-02-04 thusloop

题意:交换相邻位子，使得s字典序比t小
思路：将本位换成比原来为小的数值或将本位换成与原来位相同的数值然后继续走下去 （平衡树维护交换的位子）

//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#include<ext/rope>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int inf=8e18;
const int maxn=2e5+100;
using namespace __gnu_pbds;
tree<int, null_type, greater<int>, rb_tree_tag, tree_order_statistics_node_update> t;
int js(int x)

	return x+t.order_of_key(x);

signed main()

	IOS
	int tt;
	cin>>tt;
	while(tt--)
	
		t.clear();
		int n;
		string a,b;
		set<int>s[30];
		cin>>n;
		cin>>a>>b;
		for(int i=0; i<n; i++)
		
			s[a[i]-'a'].insert(i);
		
		int ans=inf,cnt=0;
		for(int i=0; i<n; i++)
		
			int ch=b[i]-'a';
			bool fg=0;
			for(int j=0; j<ch; j++)
			
				if(!s[j].empty())
				
					ans=min(ans,cnt+js(*s[j].begin())-i);
				
			
			if(!s[ch].empty())
			
				fg=1;
				int x=*s[ch].begin();
				if(js(x)!=i)t.insert(x);cnt+=js(x)-i;
				s[ch].erase(s[ch].begin());
				
			
			if(fg==0)break;
		
		if(ans==inf)ans=-1;
		cout<<ans<<"\\n";