poj棋盘分割（记忆化）

http://poj.org/problem?id=1191

黑书上P116 想了挺久 没想出来 想推出一公式来着 退不出来。。

想偏了  正解：递归

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<stdlib.h>

 5 #include<algorithm>

 6 #include<cmath>

 7 using namespace std;

 8 #define INF 0xfffffff

 9 #define LL long long

10 int dp[20][10][10][10][10];

11 int aa[10][10],n;

12 int divide(int k,int a,int b,int c,int d)

13 {

14     int i,j,minz = INF,g;

15     if(dp[k][a][b][c][d])

16     return dp[k][a][b][c][d];

17     if(k==n)

18     {

19         int s = 0;

20         for(i = a; i <= c ; i++)

21             for(j = b ; j <= d ; j++)

22             s+=aa[i][j];

23         dp[k][a][b][c][d] = s*s;

24         return s*s;

25     }

26     for(i = a+1 ; i <= c ; i++)

27     {

28         int s = 0;

29         for(g = i ; g <= c ; g++)

30         for(j = b ; j <= d ; j++)

31         s+=aa[g][j];

32         minz = min(minz,s*s+divide(k+1,a,b,i-1,d));

33         s = 0;

34         for(g = a ; g <= i-1 ; g++)

35         for(j = b ; j <= d ; j++)

36         s+=aa[g][j];

37         minz = min(minz,s*s+divide(k+1,i,b,c,d));

38     }

39     for(i = b+1 ; i <= d ; i++)

40     {

41         int s = 0;

42         for(g = a ; g <= c ; g++)

43         for(j = i ; j <= d ; j++)

44         s+=aa[g][j];

45         minz = min(minz,s*s+divide(k+1,a,b,c,i-1));

46         s = 0;

47         for(g = a ; g <= c ; g++)

48         for(j = b ; j <= i-1 ; j++)

49         s+=aa[g][j];

50         minz = min(minz,s*s+divide(k+1,a,i,c,d));

51     }

52     dp[k][a][b][c][d] = minz;

53     return minz;

54 }

55 int main()

56 {

57     int i,j;

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

59     {

60         double s=0;

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

62         int m = 8;

63         n--;

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

65             for(j = 1; j <= m ; j++)

66             {

67                 scanf("%d",&aa[i][j]);

68                 s+=double(aa[i][j]);

69             }

70         s = s/double(n+1);

71         int ans = divide(0,1,1,m,m);

72         double an = 1.0/double(n+1)*double(ans)-s*s;

73         printf("%.3lf\n",sqrt(an));

74     }

75     return 0;

76 }

View Code

下面那个不是 贴错了

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<stdlib.h>

 5 #define N 5010

 6 #define M 3010

 7 #define INF 0xfffffff

 8 using namespace std;

 9 int dp[M][N],h[N];

10 int main()

11 {

12     int i,j,n,m,g;

13     scanf("%d%d",&m,&n);

14     for(i = 1; i <= n ;i++)

15     scanf("%d",&h[i]);

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

17         for(j = 1; j <= n ;j++)

18         dp[i][j] = INF;

19     int o = INF;

20     for(i = 2 ; i < n ;i++)

21     {

22         dp[1][i] = min(o,(h[i]-h[i-1])*(h[i]-h[i-1]));

23         o = min(dp[1][i],o);

24     }

25     for(i = 2 ; i <= m ; i++)

26     {

27         int o = INF;

28         for(j = 2*i ; j < n-(m-i)*3 ;j++)

29         {

30             if(j-2<(n-(m-i+1)*3))

31             dp[i][j] = min(o,dp[i-1][j-2]+(h[j]-h[j-1])*(h[j]-h[j-1]));

32             else

33             dp[i][j] = min(o,dp[i-1][j-3]+(h[j]-h[j-1])*(h[j]-h[j-1]));

34             o = min(o,dp[i][j]);

35         }

36     }

37     printf("%d\n",dp[m][n-1]);

38     return 0;

39 }

View Code