B. New Year and Ascent Sequence (+ sorted binary)

https://codeforces.ml/problemset/problem/1284/B

The meaning of problems : you n sequences , two sequences are optionally concatenated, the sequence p and q linking sequence, p + q make AI. <AJ ( . 1 <= I <J <= n ). You have to ask how many connections.

Method 1: If the sequence number itself satisfies a condition ans , the contribution of the sequence of 2 * ans * (n-ans ) + ans * (ans + 1) - ans.

Sequence of maximum and minimum values of the statistical condition is not satisfied, the maximum value of sorting through the array, if the minimum value. <Maximum value of the condition is met, contributions : to the current position of the maximum number of how many have accumulated up to the end.

Solution 2: n is anti difficult, statistics ineligible number, minimum and maximum decrement if the recording sequence, the recording (small maximum minimum order of another sequence) decreasing the number of the combined sequence.

#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e7+9;
const int maxn = 1e5+9;
const double esp = 1e-6;
vector<pii>pr;


void solve(){
    int n ;
    cin >> n ;
    int ans = n * n ;
    rep(i , 1 , n){
        int m ; cin >> m;
        vector<int>v(m);
        rep(i , 0 , m-1) cin >> v[i];
        reverse(all(v));
        if(is_sorted(all(v))){
            pr.emplace_back(v[0] , v.back());
        }
    }
    sort(all(pr));
    rep(i , 0 , size(pr)-1){
            ans -= pr.end() - lower_bound(all(pr) , pii(pr[i].se , -1));
    }
        cout << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    //int t;
    //scanf("%lld" , &t);
    //while(t--)
    //while(~scanf("%lld%lld" , &n , &m) && (n || m))
        solve();
}