1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 int dp[32][32][12];
 7 bool mole[32][32][12];
 8 int n, d, m, maxt, mint;
 9 int online(int t,int x1,int y1,int x2,int y2)
10 {
11     int sx = min(x1, x2), ex = max(x1, x2);
12     int sy = min(y1, y2), ey = max(y1, y2);
13     int res = 0;
14     for (int x = sx; x <= ex; x++)
15         for (int y = sy; y <= ey; y++)
16             if (mole[x][y][t] && (y1 - y2) * (x - x2) == (y - y2) * (x1 - x2))
17                 res++;
18     return res;
19 }
20 bool judge(int x1,int y1,int x2,int y2)//判断距离是否不大于d
21 {
22     int tem = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
23     return tem <= d * d;
24 }
25 int getdp()
26 {
27     memset(dp, 0, sizeof(dp));
28     int res = 0;
29     for (int i = mint; i <= maxt; i++)//枚举时间点
30         for (int x = 0; x < n; x++)
31             for (int y = 0; y <= n; y++)//枚举锤子落点
32             {
33                 int sx = max(x - d, 1), ex = min(x + d, n);
34                 int sy = max(y - d, 1), ey = min(y + d, n);
35                 for (int px = sx; px <= ex; px++)
36                     for (int py = sy; py <= ey; py++) if (judge(x, y, px, py))//枚举锤子上次的合法落点
37                     {
38                         dp[x][y][i+1] = max(dp[x][y][i+1], dp[px][py][i] + online(i, x, y, px, py));
39                         res = max(res, dp[x][y][i+1]);
40                     }
41             }
42     return res;
43 }
44 int main()
45 {
46     //freopen("data.in", "r", stdin);
47     while (~scanf("%d%d%d", &n, &d, &m) && (n | d | m))
48     {
49         memset(mole, 0, sizeof(mole));
50         n += 10;
51         maxt = 0;
52         mint = 0x3f3f3f3f;
53         for (int i = 1; i <= m; i++)
54         {
55             int x, y, t;
56             scanf("%d%d%d",&x, &y, &t);
57             mole[x+5][y+5][t] = 1;
58             maxt = max(maxt, t);
59             mint = min(mint, t);
60         }
61         printf("%d\n",getdp());
62     }
63     return 0;
64 }