bzoj 4000 矩阵快速幂优化DP

 

建立矩阵,跑快速幂

 

 1 /**************************************************************
 2     Problem: 4000
 3     User: idy002
 4     Language: C++
 5     Result: Accepted
 6     Time:32 ms
 7     Memory:836 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12 #define M 6
13  
14 struct Matrix {
15     unsigned v[1<<M][1<<M];
16     const unsigned *operator[]( int i ) const { return v[i]; }
17 };
18  
19 int n, m, p, k, bound;
20 int attack[3];
21 int stat[1<<M], id[1<<M], stot;
22 Matrix base, dest;
23  
24 void make_unit( Matrix &x ) {
25     for( int i=0; i<stot; i++ )
26         for( int j=0; j<stot; j++ )
27             x.v[i][j] = i==j;
28 }
29 Matrix operator*( const Matrix &a, const Matrix &b ) {
30     Matrix c;
31     for( int i=0; i<stot; i++ )
32         for( int j=0; j<stot; j++ ) {
33             c.v[i][j] = 0;
34             for( int k=0; k<stot; k++ )
35                 c.v[i][j] += a[i][k]*b[k][j];
36         }
37     return c;
38 }
39 Matrix mpow( Matrix a, int b ) {
40     Matrix rt;
41     for( make_unit(rt); b; b>>=1,a=a*a ) 
42         if( b&1 ) rt=rt*a;
43     return rt;
44 }
45 int getarea( int s, int a ) {
46     int rt = 0;
47     for( int b=0; b<m; b++ ) {
48         if( (s>>b)&1 ) {
49             int aa = a;
50             if( b<k ) 
51                 aa >>= k-b;
52             else
53                 aa <<= b-k;
54             aa &= bound;
55             rt |= aa;
56         }
57     }
58     return rt;
59 }
60 void build() {
61     stot = 0;
62     memset( id, -1, sizeof(id) );
63     for( int s=0; s<=bound; s++ ) {
64         if( getarea(s,attack[1])&s ) continue;
65         stat[stot]=s;
66         id[s] = stot;
67         stot++;
68     }
69     for( int s1=0; s1<=bound; s1++ ) {
70         if( id[s1]==-1 ) continue;
71         for( int s2=0; s2<=bound; s2++ ) {
72             if( id[s2]==-1 ) continue;
73             if( getarea(s1,attack[2])&s2 ) continue;
74             if( getarea(s2,attack[0])&s1 ) continue;
75             base.v[id[s1]][id[s2]] = 1;
76         }
77     }
78 }
79 int main() {
80     scanf( "%d%d%d%d", &n, &m, &p, &k );
81     bound = (1<<m)-1;
82     for( int i=0; i<3; i++ )
83         for( int j=0,o; j<p; j++ ) {
84             scanf( "%d", &o );
85             attack[i] = attack[i] | (o<<j);
86         }
87     attack[1] ^= 1<<k;
88     build();
89     dest = mpow( base, n-1 );
90     unsigned ans = 0;
91     for( int i=0; i<stot; i++ )
92         for( int j=0; j<stot; j++ )
93             ans += dest[i][j];
94     printf( "%u\n", ans );
95 }
View Code

 

posted @ 2015-06-09 17:26  idy002  阅读(177)  评论(0编辑  收藏  举报