Codeforces Round #310 (Div. 1)

A. Case of Matryoshkas

只要统计和1相连的有多少个就好了(1,2,3..k),不相连的都要断开，剩下的链，每条链都要断开，代价是第i条链的长度-1，最后再合起来，代价是所有节点个数-1

数组开成了10000 FST了

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 using namespace std;
 6
 7 typedef long long ll;
 8
 9 int A[100000 + 5];
10
11 int main() {
12     int n, k, tp, m;
13     int ans = 0, cnt = 0;
14     scanf("%d%d", &n, &k);
15     for (int i = 1; i <= k; i++) {
16         scanf("%d", &m);
17         bool flag = 0;
18         for (int j = 0; j < m; j++) {
19             scanf("%d", &tp);
20             if (tp == 1) {
21                 flag = 1;
22                 A[1] = 1;
23                 for (int k = 2; k <= m; k++) {
24                     scanf("%d", &A[k]);
25                 }
26                 for (int k = 1; k <= m; k++) {
27                     if (A[k] == k) {
28                         cnt++;
29                     } else {
30                         break;
31                     }
32                 }
33                 ans += m-cnt;
34                 break;
35             }
36         }
37         if (!flag) ans += m-1;
38     }
39     ans += n-cnt;
40     printf("%d\n", ans);
41     return 0;
42 }