BZOJ4000 [TJOI2015]棋盘

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！

 /**************************************************************
     Problem: 4000
     User: rausen
     Language: C++
     Result: Accepted
     Time:208 ms
     Memory:920 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <vector>

 using namespace std;
 typedef unsigned int uint;
 const int N = ;

 inline int read();

 int n, m, p, K, Mx;
 int a[N][N], f[N];
 vector <int> mp[];
 uint ans;

 struct mat {
     uint x[][];
     mat() {
         memset(x, , sizeof(x));
     }
     inline void one() {
         static int i;
         *this = mat();
         for (i = ; i < Mx; ++i) x[i][i] = ;
     }
     inline uint* operator [] (int t) {
         return x[t];
     }

     inline mat operator * (const mat &m) const {
         static mat res;
         static int i, j, k;
         res = mat();
         for (i = ; i < Mx; ++i)
             for (j = ; j < Mx; ++j)
                 for (k = ; k < Mx; ++k)
                     res[i][j] += x[i][k] * m.x[k][j];
         return res;
     }
 } A;

 inline mat pow(const mat& m, int y) {
     static mat x, res;
     x = m, res.one();
     int i, j;
     while (y) {
         if (y & ) res = res * x;
         x = x * x, y >>= ;
     }
     return res;
 }

 inline bool in(int x) {
     return  <= x && x < m;
 }

 inline bool check(int s1, int s2) {
     static int i, j, t;
     for (i = ; i < m; ++i) if ((s1 >> i) & ) {
         for (j = ; j < mp[].size(); ++j)
             if (in(t = i + mp[][j]) && ((s1 >> t) & )) return ;
         for (j = ; j < mp[].size(); ++j)
             if (in(t = i + mp[][j]) && ((s2 >> t) & )) return ;
     }
     for (i = ; i < m; ++i) if ((s2 >> i) & ) {
         for (j = ; j < mp[].size(); ++j)
             if (in(t = i + mp[][j]) && ((s2 >> t) & )) return ;
         for (j = ; j < mp[].size(); ++j)
             if (in(t = i + mp[][j]) && ((s1 >> t) & )) return ;
     }
     return ;
 }

 int main() {
     int i, j;
     n = read(), m = read(), p = read(), K = read(), Mx =  << m;
     for (i = ; i < ; ++i)
         for (j = ; j < p; ++j)
             if (read() && !(i ==  && j == K)) mp[i].push_back(j - K);
     for (i = ; i < Mx; ++i)
         for (j = ; j < Mx; ++j) A[i][j] = check(i, j);
     for (i = ; i < Mx; ++i) f[i] = bool(A[][i]);
     A = pow(A, n);
     for (i = ; i < Mx; ++i) ans += f[i] * A[i][];
     printf("%u\n", ans);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }

(p.s. 这道题是0开始标号的。。。注意了QAQQQ)