poj3034Whac-a-Mole(dp)

链接

状态转移好想 不过有坑 大家都犯的错误 我也会犯 很正常 

就是锤子可以移到n*n以外  要命的是我只加了5 以为最多不会超过5  WA了N久  才想到 上下两方向都可以到5 所以最多加10

以时间和坐标进行DP

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<cmath>

 7 using namespace std;

 8 #define N 1010

 9 #define M 32

10 struct node

11 {

12     int x[N],y[N];

13 }f[M][M];

14 int o[12][M][M];

15 int dp[12][M][M];

16 int n,nu[M][M];

17 int dis(int x1,int y1,int x2,int y2)

18 {

19     if(x1<0||x1>=n+10||y1<0||y1>=n+10)

20     return -1;

21     int s;

22     s = (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);

23     return s;

24 }

25 int main()

26 {

27     int i,j,g,k,d,kk,m;

28     while(scanf("%d%d%d",&n,&d,&m)!=EOF)

29     {

30         if(n==0&&d==0&&m==0)

31         break;

32         memset(dp,0,sizeof(dp));

33         memset(o,0,sizeof(o));

34         memset(nu,0,sizeof(nu));

35         int maxz = 0;

36         int a,b,c;

37         for(i = 1; i <= m ; i++)

38         {

39             scanf("%d%d%d",&a,&b,&c);

40             o[c][a+5][b+5] =1;

41             maxz = max(maxz,c);

42         }

43         int ans=0;

44         for(i = 0 ; i < n+10 ; i++)

45             for(j = 0 ;j < n+10 ; j++)

46             {

47                  int po = 0;

48                  for(k = -d; k <= d; k++)

49                     for(kk = -d; kk <= d ; kk++)

50                     {

51                         int tx = i+k;

52                         int ty = j+kk;

53                         if(dis(tx,ty,i,j)!=-1&&dis(tx,ty,i,j)<=d*d)

54                         {

55                             po++;

56                             f[i][j].x[po] = tx;

57                             f[i][j].y[po] = ty;

58                             nu[i][j] = po;

59                         }

60                     }

61             }

62         for(i = 1 ;i <= maxz ; i++)

63             for(j = 0; j < n+10 ; j++)

64                 for(g = 0 ; g < n+10 ; g++)

65                    for(k = 1; k <= nu[j][g] ; k++)

66                    {

67                        int tx = f[j][g].x[k];

68                        int ty = f[j][g].y[k];

69                        int x1 = min(tx,j),x2 = max(tx,j);

70                        int y1 = min(ty,g),y2 = max(ty,g);

71                        int num = 0;

72                        for(kk = x1; kk <= x2 ; kk++)

73                         for(int mm = y1 ; mm <= y2 ; mm++)

74                         {

75                             if((ty-g)*(kk-j)==(tx-j)*(mm-g)&&o[i][kk][mm])

76                             num++;

77                         }

78                         dp[i][tx][ty] = max(dp[i][tx][ty],dp[i-1][j][g]+num);

79                         ans = max(ans,dp[i][tx][ty]);

80                     }

81         printf("%d\n",ans);

82     }

83     return 0;

84 }

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