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 <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 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<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef pair<LL, LL> pll; 65 typedef map<string, int> msi; 66 typedef vector<int> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 const int maxn = 101; 72 const int maxm = 11; 73 char tmp[maxm]; 74 int dp[1<<maxm][1<<maxm][maxn]; 75 int G[maxn]; 76 int n, m; 77 78 bool ok(int x, int i) { 79 if((x & G[i]) != x) return 0; 80 if((x << 1) & x != 0) return 0; 81 if((x << 2) & x != 0) return 0; 82 return 1; 83 } 84 85 signed main() { 86 // FRead(); 87 while(~scanf("%d%d",&n,&m)) { 88 Cls(G); 89 int mm = 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 ", ret); 124 } 125 RT 0; 126 }