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

 

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 1 /*  2  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  3  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，  4  其他是2次操作，还要减去k-1个娃娃是只要套上就可以  5  详细解释：http://blog.csdn.net/firstlucker/article/details/46671251  6 */  7 #include <cstdio>  8 #include <algorithm>  9 #include <cstring> 10 #include <cmath> 11 using namespace std; 12 13 const int MAXN = 1e5 + 10; 14 const int INF = 0x3f3f3f3f; 15 int a[MAXN]; 16 17 int main(void) //Codeforces Round #310 (Div. 2) C. Case of Matryoshkas 18 { 19 // freopen ("C.in", "r", stdin); 20 21 int n, k; 22 while (scanf ("%d%d", &n, &k) == 2) 23  { 24 int cnt = 0; 25 for (int i=1; i<=k; ++i) 26  { 27 int m; scanf ("%d", &m); int p = 0; 28 for (int j=1; j<=m; ++j) 29  { 30 scanf ("%d", &a[j]); if (a[j] == 1) p = j; 31  } 32 if (p) 33  { 34 for (int j=p+1; j<=m; ++j) 35  { 36 if (a[j] == a[j-1] + 1) cnt++; 37 else break; 38  } 39  } 40  } 41 printf ("%d\n", (n - 1 - cnt) * 2 - (k - 1)); 42  } 43 44 return 0; 45 }