Pebbles

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 777 Accepted Submission(s): 426

Problem Description

You're given an unlimited number of pebbles to distribute across an N x N game board (N drawn from [3, 15]), where each square on the board contains some positive point value between 10 and 99, inclusive. A 6 x 6 board might look like this:

The player distributes pebbles across the board so that:

?At most one pebble resides in any given square.

?No two pebbles are placed on adjacent squares. Two squares are considered adjacent if they are horizontal, vertical, or even diagonal neighbors. There's no board wrap, so 44 and 61 of row three aren't neighbors. Neither are 33 and 75 nor 55 and 92.

The goal is to maximize the number of points claimed by your placement of pebbles.

Write a program that reads in a sequence of boards from an input file and prints to stdout the maximum number of points attainable by an optimal pebble placement for each.

Input

Each board is expressed as a series of lines, where each line is a space-delimited series of numbers. A blank line marks the end of each board (including the last one)

Output

then your program would print the maximum number of points one can get by optimally distributing pebbles while respecting the two rules, which would be this (each output should be printed on a single line and followed with a newline):

Sample Input

71 24 95 56 54

85 50 74 94 28

92 96 23 71 10

23 61 31 30 46

64 33 32 95 89

78 78 11 55 20 11

98 54 81 43 39 97

12 15 79 99 58 10

13 79 83 65 34 17

85 59 61 12 58 97

40 63 97 85 66 90

33 49 78 79 30 16 34 88 54 39 26

80 21 32 71 89 63 39 52 90 14 89

49 66 33 19 45 61 31 29 84 98 58

36 53 35 33 88 90 19 23 76 23 76

77 27 25 42 70 36 35 91 17 79 43

33 85 33 59 47 46 63 75 98 96 55

75 88 10 57 85 71 34 10 59 84 45

29 34 43 46 75 28 47 63 48 16 19

62 57 91 85 89 70 80 30 19 38 14

61 35 36 20 38 18 89 64 63 88 83

45 46 89 53 83 59 48 45 87 98 21

15 95 24 35 79 35 55 66 91 95 86 87

94 15 84 42 88 83 64 50 22 99 13 32

85 12 43 39 41 23 35 97 54 98 18 85

84 61 77 96 49 38 75 95 16 71 22 14

18 72 97 94 43 18 59 78 33 80 68 59

26 94 78 87 78 92 59 83 26 88 91 91

34 84 53 98 83 49 60 11 55 17 51 75

29 80 14 79 15 18 94 39 69 24 93 41

66 64 88 82 21 56 16 41 57 74 51 79

49 15 59 21 37 27 78 41 38 82 19 62

54 91 47 29 38 67 52 92 81 99 11 27

31 62 32 97 42 93 43 79 88 44 54 48

Sample Output

572

683

2096

2755

Source

2007 ACM-ICPC Pacific Northwest Region

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lcy

1 //484MS    2808K    1941 B    C++      2 /* 3      4     题意： 5         给出n*n个数，求最大的不相邻的数的和(对角也算)   6      7     状态压缩：  8         dp[i][j]表示前i行中状态j的最优策略  9         这题和 hdu1565 类似，也是方格取数问题，我采用了一样的解法，10     唯一的小变形是在判断相邻行剪得判断，要加左移和右移的判断(判断对角)，11     其他和 hdu1565 差不多 12 13 */14 #include
     
      15 #include
      
       16 int dp[20][1<<15];17 int g[20][20];18 int s[2000];19 char c[100];20 int Max(int a,int b)21 {22     return a>b?a:b;23 }24 int judge(int x) //判断 25 {26     int end=0;27     while(x){28         if(end && (1&x)) return 0;29         end=1&x;30         x>>=1;31     }32     return 1;33 }34 void init() //预处理符合条件的状态 35 {36     memset(s,0,sizeof(s));37     int k=0;38     for(int i=0;i<(1<<15);i++)39         if(judge(i)) s[k++]=i;40     //printf("%d\n",k);  //k=159741 }42 int main(void)43 {44     init();45     while(gets(c)) 46     {47         int len=strlen(c);48         int n=0,t=0;49         for(int i=0;i
       
        ='0' && c[i]<='9'){52                 t=t*10+c[i]-'0';53             }else{54                 g[0][n++]=t;t=0;55             }56         }57         g[0][n++]=t;58         memset(dp,0,sizeof(dp));59         int n0=1<
        
         >1)))) //小变形，主要是处理对角情况 78                         dp[i][j]=Max(dp[i][j],temp+dp[i-1][k]);79                 maxn=Max(maxn,dp[i][j]);80             }81         }82         printf("%d\n",maxn);83         c[0]=getchar();  //后面两个用于处理换行符和空行 84         gets(c);     85     }86     return 0;87 }