HDU 2280 状压DP

用dfs找到状态的最优解

且那个正方形块可以由两个水平块组成，所以无需考虑

 

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 

 5 using namespace std;

 6 #define N 1005

 7 int state[N] , n , m , dp[N][1<<6];

 8 

 9 void dfs(int i , int k , int state1 , int state2 , int v)

10 {

11     if(k > 5){

12         dp[i][state2] = max(dp[i][state2] , v);

13         return;

14     }

15     if(k <= 4){

16         if(!(state1&(1<<k-1)) && !(state2&(3<<k-1))){

17             dfs(i , k+1 , state1 | (1<<k-1) , state2 | (3<<k-1) , v+3);

18         }

19         if(!(state1&(1<<k)) && !(state2&(3<<k-1))){

20             dfs(i , k+1 , state1 | (1<<k) , state2 | (3<<k-1) , v+3);

21         }

22         if(!(state1&(3<<k-1)) && !(state2&(1<<k-1))){

23             dfs(i , k+1 , state1 | (3<<k-1) , state2 | (1<<k-1) , v+3);

24         }

25         if(!(state1&(3<<k-1)) && !(state2&(1<<k))){

26             dfs(i , k+1 , state1 | (3<<k-1) , state2 | (1<<k) , v+3);

27         }

28         if(!(state2&(3<<k-1))){

29             dfs(i , k+1 , state1 , state2 | (3<<k-1) , v+2);

30         }

31     }

32     else{

33         if(!(state1&(1<<k-1)) && !(state2&(1<<k-1))){

34             dfs(i , k+1 , state1 | (1<<k-1) , state2 | (1<<k-1) , v+2);

35         }

36     }

37     dfs(i , k+1 , state1 , state2 , v);

38 }

39 

40 void dfs2(int k , int sta , int v)

41 {

42     if(k > 5){

43         dp[1][sta] = max(dp[1][sta] , v);

44         return;

45     }

46     if(k <= 4){

47         if(!(sta & (3<<k-1)))

48             dfs2(k+1 , sta | (3<<k-1) , v+2);

49     }

50     dfs2(k+1 , sta , v);

51 }

52 

53 int main()

54 {

55   //  freopen("a.in" , "rb" , stdin);

56     char a;

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

58         memset(state , 0 ,sizeof(state));

59         int all = 0;

60         for(int i = 1; i <= n ; i++)

61             for(int j=1; j<=5 ; j++){

62                 cin>>a;

63                 if(a == '1'){

64                     state[i] |= (1<<j-1);

65                 }

66                 else all++;

67             }

68 

69         memset(dp , -1 , sizeof(dp));

70         dp[1][state[1]] = 0;

71         dfs2(1 , state[1] , 0);

72 

73         for(int i=2 ; i <= n ; i++){

74             for(int j = 0; j<(1<<5) ; j++){

75                 if(dp[i-1][j] >= 0){

76                     dfs(i , 1 , j , state[i] , dp[i-1][j]);

77                 }

78             }

79         }

80 

81         int maxn = 0;

82 

83         for(int i = 0 ; i<(1<<5) ; i++){

84             maxn = max(maxn , dp[n][i]);

85         }

86 

87         int remain = all - maxn;

88         if(remain <= m) puts("YES");

89         else puts("NO");

90     }

91     return 0;

92 }