贪心/思维题 Codeforces Round #310 (Div. 2) C. Case of Matryoshkas

 

题目传送门

/*
    题意：套娃娃，可以套一个单独的娃娃，或者把最后面的娃娃取出，最后使得0-1-2-...-(n-1)，问最少要几步
    贪心/思维题：娃娃的状态：取出+套上(2)，套上(1), 已套上(0)，先从1开始找到已经套好的娃娃层数，
                其他是2次操作，还要减去k-1个娃娃是只要套上就可以
    详细解释：http://blog.csdn.net/firstlucker/article/details/46671251
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a[MAXN];

int main(void)      //Codeforces Round #310 (Div. 2) C. Case of Matryoshkas
{
    // freopen ("C.in", "r", stdin);

    int n, k;
    while (scanf ("%d%d", &n, &k) == 2)
    {
        int cnt = 0;
        for (int i=1; i<=k; ++i)
        {
            int m;    scanf ("%d", &m);    int p = 0;
            for (int j=1; j<=m; ++j)
            {
                scanf ("%d", &a[j]);    if (a[j] == 1)    p = j;
            }
            if (p)
            {
                for (int j=p+1; j<=m; ++j)
                {
                    if (a[j] == a[j-1] + 1)    cnt++;
                    else    break;
                }
            }
        }
        printf ("%d\n", (n - 1 - cnt) * 2 - (k - 1));
    }

    return 0;
}