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 }