找结论——势能

• https://atcoder.jp/contests/arc120/tasks/arc120_c

势能就是i + a[i],很有用的结论
对于能完成交换，从A变成B，势能数组一定相同
找出最小交换次数，贪心的方案是：最近的先交换
找到最近的在可以用数组数组和set实现
具体细节见代码

#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define debug0(x) cout << "debug0: " << x << endl
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define YES cout<<"Yes"<<endl
#define NO cout<<"No"<<endl
#define fi first
#define se second
#define int long long
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)

const int N = 2e5+10;
int aa[N];
int lowbit(int x)  // 返回末尾的1
{return x & -x;}

void add(int x,int w){
    for(int i = x;i < N;i += lowbit(i)){
        aa[i] += w;
    }
}

int query(int u){
    int res = 0;
    for(int i = u;i > 0;i -= lowbit(i)){
        res += aa[i];
    }
    return res;
}

int a[N],b[N];
multiset<int> A,B;

void solve() 
{
    int n;cin >> n;
    
    for(int i = 1; i <= n; i++)cin >> a[i],A.insert(a[i] + i);
    for(int i = 1; i <= n; i++)cin >> b[i],B.insert(b[i] + i);

    if(A != B)
    {
        cout << -1 << endl;
        return ;
    }

    set<PII> s;
    int ans = 0;
    for(int i = 1; i <= n; i++)s.insert(PII{a[i] + i, i});
    for(int i = 1; i <= n; i++)
    {
        set<PII>::iterator it = s.lower_bound(PII{b[i]+i,0});

        int t = query((*it).second) + (*it).second;
        
        add(1,1);
        assert((*it).second>0);
        add((*it).second,-1);//这里相对位置不用加1，因为这个是相对于a数组原始位置而言的
        ans += t - i;
              
        s.erase(it);
    }

    cout << ans << endl;

}

signed main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    

    int T = 1;//cin >> T;
    while(T--){
        solve();
    }
    return 0;
}