poj3034 dp

 1 //Accepted    260 KB    579 ms
 2 //题目中给出moles不会出现在负坐标上，但是没说hammer不会出现在负坐标上（简直就是扯淡）
 3 //考虑到hammer最大的移动距离是d，把坐标都右移d，剩下的就是dp了
 4 //dp[i][j][t]表示t时间hammer在（x,y）处能够得到的最大值
 5 //dp[i][j][t]=max(dp[x][y][t-1]+count)
 6 //其中(i,j)和(x,y)之间的距离小于等于d
 7 //count表示两点之间在t时刻mole的数量
 8 #include <cstdio>
 9 #include <cstring>
10 #include <iostream>
11 using namespace std;
12 const int imax_n = 32;
13 const int imax_t = 12;
14 int appearance[imax_n][imax_n][imax_t];
15 int dp[imax_n][imax_n][imax_t];
16 int n,m,t,d;
17 int max_t;
18 int max(int a,int b)
19 {
20     return a>b?a:b;
21 }
22 int min(int a,int b)
23 {
24     return a>b?b:a;
25 }
26 void Dp()
27 {
28     memset(dp,0,sizeof(dp));
29     for (int t=0;t<=max_t+1;t++)
30     {
31         for (int i=0;i<n+2*d;i++)    //考虑到hammer可以移动到moles出现以外的地方，dp的范围扩大2*d
32         {
33             for (int j=0;j<n+2*d;j++)
34             {
35                 dp[i][j][t]=dp[i][j][t-1];
36                 for (int x1=i-d;x1<=i+d;x1++)
37                 {
38                     if (x1<0) continue;
39                     if (x1>n+2*d) break;
40                     for (int y1=j-d;y1<=j+d;y1++)
41                     {
42                         if (y1<0) continue;
43                         if (y1>n+2*d) break;
44                         if ((i-x1)*(i-x1)+(j-y1)*(j-y1)>d*d) continue;
45                         int start_x,start_y,end_x,end_y;
46                         start_x=min(i,x1);
47                         end_x=max(i,x1);
48                         start_y=min(j,y1);
49                         end_y=max(j,y1);
50                         int count=0;
51                         for (int x2=start_x;x2<=end_x;x2++)
52                         for (int y2=start_y;y2<=end_y;y2++)
53                         if ((x1-i)*(y2-j)==(x2-i)*(y1-j))
54                         count+=appearance[x2][y2][t];
55                         dp[i][j][t]=max(dp[i][j][t],dp[x1][y1][t-1]+count);
56                     }
57                 }
58             }
59         }
60     }
61     int ans=0;
62     for (int i=0;i<n+2*d;i++)
63     for (int j=0;j<n+2*d;j++)
64     ans=max(ans,dp[i][j][max_t+1]);
65     printf("%d\n",ans);
66 }
67 int main()
68 {
69     while (scanf("%d%d%d",&n,&d,&m),n+m+d)
70     {
71         int x,y,t;
72         max_t=0;
73         memset(appearance,0,sizeof(appearance));
74         for (int i=0;i<m;i++)
75         {
76             scanf("%d%d%d",&x,&y,&t);
77             appearance[x+d][y+d][t]=1;
78             max_t=max(max_t,t);
79         }
80         Dp();
81     }
82     return 0;
83 }