【35.37%】【codeforces 556C】Case of Matryoshkas

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of contemporary art.

The main exhibit is a construction of n matryoshka dolls that can be nested one into another. The matryoshka dolls are numbered from 1 to n. A matryoshka with a smaller number can be nested in a matryoshka with a higher number, two matryoshkas can not be directly nested in the same doll, but there may be chain nestings, for example, 1 → 2 → 4 → 5.

In one second, you can perform one of the two following operations:

Having a matryoshka a that isn’t nested in any other matryoshka and a matryoshka b, such that b doesn’t contain any other matryoshka and is not nested in any other matryoshka, you may put a in b;

Having a matryoshka a directly contained in matryoshka b, such that b is not nested in any other matryoshka, you may get a out of b.

According to the modern aesthetic norms the matryoshka dolls on display were assembled in a specific configuration, i.e. as several separate chains of nested matryoshkas, but the criminal, following the mysterious plan, took out all the dolls and assembled them into a single large chain (1 → 2 → … → n). In order to continue the investigation Andrewid needs to know in what minimum time it is possible to perform this action.

Input

The first line contains integers n (1 ≤ n ≤ 105) and k (1 ≤ k ≤ 105) — the number of matryoshkas and matryoshka chains in the initial configuration.

The next k lines contain the descriptions of the chains: the i-th line first contains number mi (1 ≤ mi ≤ n), and then mi numbers ai1, ai2, …, aimi — the numbers of matryoshkas in the chain (matryoshka ai1 is nested into matryoshka ai2, that is nested into matryoshka ai3, and so on till the matryoshka aimi that isn’t nested into any other matryoshka).

It is guaranteed that m1 + m2 + … + mk = n, the numbers of matryoshkas in all the chains are distinct, in each chain the numbers of matryoshkas follow in the ascending order.

Output

In the single line print the minimum number of seconds needed to assemble one large chain from the initial configuration.

Examples

input

3 2

2 1 2

1 3

output

1

input

7 3

3 1 3 7

2 2 5

2 4 6

output

10

Note

In the first sample test there are two chains: 1 → 2 and 3. In one second you can nest the first chain into the second one and get 1 → 2 → 3.

In the second sample test you need to disassemble all the three chains into individual matryoshkas in 2 + 1 + 1 = 4 seconds and then assemble one big chain in 6 seconds.

【题目链接】:http://codeforces.com/contest/556/problem/C

【题解】



英文题………

这个是俄罗斯套娃。。。。。。



读懂题意之后..

其实很简单了。。

每个套娃必须是单个才能套在其他套娃身上..(其他套娃里面可以套没关系,但是它本身必须是在最外面的);

且拆也只能一个一个的拆.

但是最后又必须是1..n的顺序;

则只能是序号为1的套娃,那连续的几个套娃能够节省,否则都要拆成1个一个的.

比如

7 2

4 1 2 3 5

3 4 6 7

则只有1 2 3这个可以不用拆开

剩余的都要拆成单个

即5 4 6 7都拆成单个的.

然后再一个一个地拼在1 2 3的后面.(1 2 3作为整体);

而且

7 2

4 5 1 2 3

3 4 6 7

这种也一个都不能节省需要全部都拆成单个的

(1 2 3也不能作为整体了);

因为1 2 3前面有一个5;

这个5肯定要拆开的…

而要拆开5 只能把3 、2、1都一个一个地拆开…



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
#define pri(x) printf("%d",x)
#define prl(x) printf("%I64d",x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 1e5+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,k;
int b[MAXN];
vector <int> a;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    int ans = 0,tot =0;
    rei(n);rei(k);
    rep1(i,1,k)
    {
        int num,temp1=0;
        rei(num);
        rep1(i,1,num)
            rei(b[i]);
        temp1+=num;
        int now = 1;
        if (b[now]==1)
        {
            while (now+1<=num && b[now+1]==b[now]+1)
                now++;
            temp1-=now;
        }
        tot+=temp1;
        if (temp1>0)
            ans+=temp1-1;
        if (temp1>0 && temp1 < num)
            ans++;
    }
    //cout << ans << endl;
    ans+=tot;
    cout << ans << endl;
    return 0;
}