[POJ1185]炮兵阵地（状压DP）-白红宇

[POJ1185]炮兵阵地（状压DP）

阅读量：4592 次

发布时间：2019-06-09

本文共 2739 字，大约阅读时间需要 9 分钟。

题目链接：http://poj.org/problem?id=1185

这个和之前的不一样，在于某个点影响的范围是两格。那么dp(cur,pre,i)表示第i行状态为cur，i-1行状态为pre时可以有多少种放法。转移的时候枚举ppre，就是i-2行即可。照葫芦画瓢

1 /*  2 ━━━━━┒ギリギリ♂ eye！  3 ┓┏┓┏┓┃キリキリ♂ mind！  4 ┛┗┛┗┛┃＼○／  5 ┓┏┓┏┓┃ /  6 ┛┗┛┗┛┃ノ)  7 ┓┏┓┏┓┃  8 ┛┗┛┗┛┃  9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include 
     
       18 #include 
      
        19 #include 
       
         20 #include 
        
          21 #include 
         
           22 #include 
          
            23 #include 
           
             24 #include 
            
              25 #include 
             
               26 #include 
              
                27 #include 
               
                 28 #include 
                
                  29 #include 
                 
                   30 #include 
                  
                    31 #include 
                   
                     32 #include 
                    
                      33 #include 
                      34 #include 
                      
                        35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onecnt(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair
                       
                         pii; 63 typedef pair
                        
                          psi; 64 typedef pair
                         
                           pll; 65 typedef map
                          
                            msi; 66 typedef vector
                           
                             vi; 67 typedef vector
                            
                              vl; 68 typedef vector
                             
                               vvl; 69 typedef vector
                              
                                vb; 70 71 const int maxn = 101; 72 const int maxm = 11; 73 char tmp[maxm]; 74 int dp[1<
                               
                                <
                                
                                 << m; 90 For(i, 1, n+1) { 91 Rs(tmp); 92 for(int j = m - 1; j >= 0; j--) { 93 G[i] <<= 1; 94 G[i] |= (tmp[j] == 'P' ? 1 : 0); 95 } 96 } 97 Rep(i, mm) { if(!ok(i, 1)) continue; 98 dp[i][0][1] = __builtin_popcount(i); 99 }100 Rep(i, mm) { if(!ok(i, 1)) continue;101 Rep(j, mm) { if(!ok(j, 2)) continue;102 if(i & j) continue;103 dp[i][j][2] = max(dp[i][j][2], dp[i][0][1] + __builtin_popcount(i));104 }105 }106 For(i, 3, n+1) {107 Rep(cur, mm) { if(!ok(cur, i)) continue;108 Rep(pre, mm) { if(!ok(pre, i-1)) continue;109 if(cur & pre) continue;110 Rep(ppre, mm) {111 if((cur & pre) || (cur & ppre) || (pre & ppre)) continue;112 dp[cur][pre][i] = max(dp[cur][pre][i], dp[pre][ppre][i-1] + __builtin_popcount(cur));113 }114 }115 }116 }117 int ret = 0;118 Rep(i, mm) {119 Rep(j, mm) {120 ret = max(ret, dp[i][j][n]);121 }122 }123 printf("%d\n", ret);124 }125 RT 0;126 }