百练雷涛的小猫

思路：
dp，递推关系优化。
实现：
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 int n, m, h, p, tmp, maxH;
 7 int a[2005][2005], dp[2005][2005];
 8 
 9 void solve()
10 {
11     for (int j = maxH; j >= 0; j--)
12     {
13         int tmp = 0;
14         for (int k = 0; k < n; k++)
15         {
16             if (j + h <= maxH)
17                 tmp = max(tmp, dp[k][j + h]);
18         }
19         for (int i = 0; i < n; i++)
20         {
21             dp[i][j] = tmp;
22             if (j + 1 <= maxH)
23                 dp[i][j] = max(dp[i][j], dp[i][j + 1]);
24             dp[i][j] += a[i][j];
25         }
26     }
27     int maxn = 0;
28     for (int i = 0; i < n; i++)
29     {
30         maxn = max(maxn, dp[i][0]);
31     }
32     cout << maxn << endl;
33 }
34 
35 int main()
36 {
37     cin >> n >> m >> h;
38     for (int i = 0; i < n; i++)
39     {
40         scanf("%d", &p);
41         for (int j = 0; j < p; j++)
42         {
43             scanf("%d", &tmp);
44             maxH = max(maxH, tmp);
45             a[i][tmp]++;
46         }
47     }
48     solve();
49     return 0;
50 }
百练 雷涛的小猫

百练雷涛的小猫
