[POJ1185]炮兵阵地（状压DP）

Posted 2020-08-09

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\n", ret);
124     }
125     RT 0;
126 }