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 }

 

 posted on 2014-11-02 12:24  Love风吟  阅读(198)  评论(0)  编辑  收藏  举报