首先是状态压缩DP。。。
然后我们发现转移都是一样的。。。可以矩阵优化。。。
于是做完啦QAQQQ
题目读不懂？恩多读几遍就读懂了，诶诶诶！别打我呀！
1 /************************************************************** 2 Problem: 4000 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:208 ms 7 Memory:920 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <vector> 13 14 using namespace std; 15 typedef unsigned int uint; 16 const int N = 75; 17 18 inline int read(); 19 20 int n, m, p, K, Mx; 21 int a[N][N], f[N]; 22 vector <int> mp[3]; 23 uint ans; 24 25 struct mat { 26 uint x[64][64]; 27 mat() { 28 memset(x, 0, sizeof(x)); 29 } 30 inline void one() { 31 static int i; 32 *this = mat(); 33 for (i = 0; i < Mx; ++i) x[i][i] = 1; 34 } 35 inline uint* operator [] (int t) { 36 return x[t]; 37 } 38 39 inline mat operator * (const mat &m) const { 40 static mat res; 41 static int i, j, k; 42 res = mat(); 43 for (i = 0; i < Mx; ++i) 44 for (j = 0; j < Mx; ++j) 45 for (k = 0; k < Mx; ++k) 46 res[i][j] += x[i][k] * m.x[k][j]; 47 return res; 48 } 49 } A; 50 51 inline mat pow(const mat& m, int y) { 52 static mat x, res; 53 x = m, res.one(); 54 int i, j; 55 while (y) { 56 if (y & 1) res = res * x; 57 x = x * x, y >>= 1; 58 } 59 return res; 60 } 61 62 inline bool in(int x) { 63 return 0 <= x && x < m; 64 } 65 66 inline bool check(int s1, int s2) { 67 static int i, j, t; 68 for (i = 0; i < m; ++i) if ((s1 >> i) & 1) { 69 for (j = 0; j < mp[1].size(); ++j) 70 if (in(t = i + mp[1][j]) && ((s1 >> t) & 1)) return 0; 71 for (j = 0; j < mp[2].size(); ++j) 72 if (in(t = i + mp[2][j]) && ((s2 >> t) & 1)) return 0; 73 } 74 for (i = 0; i < m; ++i) if ((s2 >> i) & 1) { 75 for (j = 0; j < mp[1].size(); ++j) 76 if (in(t = i + mp[1][j]) && ((s2 >> t) & 1)) return 0; 77 for (j = 0; j < mp[0].size(); ++j) 78 if (in(t = i + mp[0][j]) && ((s1 >> t) & 1)) return 0; 79 } 80 return 1; 81 } 82 83 int main() { 84 int i, j; 85 n = read(), m = read(), p = read(), K = read(), Mx = 1 << m; 86 for (i = 0; i < 3; ++i) 87 for (j = 0; j < p; ++j) 88 if (read() && !(i == 1 && j == K)) mp[i].push_back(j - K); 89 for (i = 0; i < Mx; ++i) 90 for (j = 0; j < Mx; ++j) A[i][j] = check(i, j); 91 for (i = 0; i < Mx; ++i) f[i] = bool(A[0][i]); 92 A = pow(A, n); 93 for (i = 0; i < Mx; ++i) ans += f[i] * A[i][0]; 94 printf("%u ", ans); 95 return 0; 96 } 97 98 inline int read() { 99 static int x; 100 static char ch; 101 x = 0, ch = getchar(); 102 while (ch < '0' || '9' < ch) 103 ch = getchar(); 104 while ('0' <= ch && ch <= '9') { 105 x = x * 10 + ch - '0'; 106 ch = getchar(); 107 } 108 return x; 109 }
(p.s. 这道题是0开始标号的。。。注意了QAQQQ)