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[POJ1185]炮兵阵地(状压DP)

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 }

 

[POJ1185]炮兵阵地(状压DP)