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

     return ;
 }