POJ 3034 Whac-a-Mole(DP)

题目链接 

理解了题意之后，这个题感觉状态转移还是挺好想的，实现起来确实有点繁琐，代码能力还有待加强，经过很长时间才发现bug。注意坐标可能是负的。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 #include <cmath>

 5 #include <algorithm>

 6 using namespace std;

 7 int dp[12][40][41];

 8 bool o[12][41][41];

 9 int n,ans,d;

10 int gcd(int a,int b)

11 {

12     return b == 0?a:gcd(b,a%b);

13 }

14 int fun(int t,int x,int y,int tx,int ty,int ax,int ay)

15 {

16     int temp = 0;

17     while(x != ax||y != ay)//开始这里写成&&

18     {

19         if(o[t+1][x][y])

20         temp ++;

21         x += tx;

22         y += ty;

23     }

24     return temp+o[t+1][x][y];

25 }

26 void judge(int t,int x,int y)

27 {

28     int i,j,g;

29     for(i = x-d;i <= x+d;i ++)

30     {

31         for(j = y-d;j <= y+d;j ++)

32         {

33            if((i-x)*(i-x)+(j-y)*(j-y) > d*d) continue;

34            if(i < 0||j < 0) continue;

35            if(i > n+10||j > n+10) continue;

36            g = gcd(abs(i-x),abs(j-y));

37            if(g == 0) g = 1;

38            int tx,ty;

39            tx = (i-x)/g;

40            ty = (j-y)/g;

41            dp[t+1][i][j] = max(dp[t+1][i][j],dp[t][x][y]+fun(t,x,y,tx,ty,i,j));

42            ans = max(dp[t+1][i][j],ans);

43         }

44     }

45 }

46 int main()

47 {

48     int m,i,j,k,y,x,t;

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

50     {

51         if(n == 0&&d == 0&&m == 0) break;

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

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

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

55         {

56             scanf("%d%d%d",&x,&y,&t);

57             o[t][x+5][y+5] = 1;

58         }

59         ans = 0;

60         for(i = 0; i <= 10; i ++)

61         {

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

63             {

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

65                 {

66                     judge(i,j,k);

67                 }

68             }

69         }

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

71     }

72     return 0;

73 }