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！

 

  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\n", 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 }
View Code

(p.s. 这道题是0开始标号的。。。注意了QAQQQ)

posted on 2015-05-12 23:12  Xs酱~  阅读(635)  评论(0)  编辑  收藏  举报