BZOJ3791 作业

首先我们发现嘛。。。最多可以搞出2 *k - 1段不同的

于是一遍扫过去dp就可以啦，需要注意滚动数组

 1 /**************************************************************
 2     Problem: 3791
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:280 ms
 7     Memory:1196 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13
14 using namespace std;
15 const int N = 100005;
16 const int K =55;
17
18 int n, k, ans;
19 int a[N], f[2][K << 1][2];
20
21 int main() {
22     int i, j, x, y, now;
23     scanf("%d%d", &n, &k);
24     for (i = 1; i <= n; ++i)
25         scanf("%d", a + i);
26     f[0][1][a[1]] = 1;
27     for (i = now = 1; i <= n - 1; ++i, now ^= 1) {
28         memset(f[now], 0, sizeof(f[now]));
29         for (j = 1; j <= k * 2 - 1; ++j)
30             for (x = 0; x <= 1; ++x)
31                 for (y = 0; y <= 1; ++y)
32                     f[now][j + (x != y)][y] = max(f[now][j + (x != y)][y], f[!now][j][x] + (a[i + 1] == y));
33         for (j = 1; j <= k * 2 - 1; ++j)
34             ans = max(ans, max(f[now][j][0], f[now][j][1]));
35     }
36     printf("%d\n", ans);
37     return 0;
38 }