HDU 1565 方格取数(1)(状态压缩DP）

题目链接

水过的，就是普普通通的枚举上一层和这一层，应该可以加上队列优化，降复杂度。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #define eps 1e-9

 6 #define INF -10000000

 7 #define ll __int64

 8 using namespace std;

 9 int dp[21][18000],p[21][21],o[18000];

10 int main()

11 {

12     int n,i,j,k,u,tem,num;

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

14     {

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

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

17         {

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

19                 scanf("%d",&p[i][j]);

20         }

21         num = 1;

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

23         {

24             if(i&(i<<1)||i&(i>>1))

25             {

26                 continue;

27             }

28             dp[1][num] = 0;

29             o[num] = i;

30             for(j = 0; j <= n-1; j ++)

31             {

32                 if(i&(1<<j))

33                 {

34                     dp[1][num] += p[1][j+1];

35                 }

36             }

37             num ++;

38         }

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

40         {

41             for(k = 1; k < num; k ++)

42             {

43                 for(j = 1; j < num; j ++)

44                 {

45                     if(o[j]&o[k])

46                         continue;

47                     tem = 0;

48                     for(u = 0; u <= n-1; u ++)

49                     {

50                         if(o[k]&(1<<u))

51                             tem += p[i][u+1];

52                     }

53                     if(dp[i][k] < dp[i-1][j]+tem)

54                         dp[i][k] = dp[i-1][j]+tem;

55                 }

56             }

57         }

58         /*for(i = 1; i <= n; i ++)

59         {

60             for(j = 1; j < num; j ++)

61             {

62                 printf("%d ",dp[i][j]);

63             }

64             printf("\n");

65         }*/

66         tem = 0;

67         for(i = 1; i < num; i ++)

68         {

69             if(tem < dp[n][i])

70                 tem = dp[n][i];

71         }

72         printf("%d\n",tem);

73     }

74     return 0;

75 }