[BZOJ4000][TJOI2015]棋盘(状压DP+矩阵快速幂)

题意极其有毒，注意给的行列都是从0开始的。

状压DP，f[i][S]表示第i行状态为S的方案数，枚举上一行的状态转移。$O(n2^{2m})$

使用矩阵加速，先构造矩阵a[S1][S2]表示上一行为S1是下一行是否能为S2，快速幂加速后得解。$O(2^{3m}m^2+2^{3m}\log n)$

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 5 typedef unsigned int ul;
 6 using namespace std;
 7 
 8 const int N=70;
 9 ul ans;
10 int n,m,p,k,ed,x,tot;
11 struct P{ int x,y; }d[110];
12 
13 struct Mat{
14     ul a[N][N];
15     ul* operator [](int x){ return a[x]; }
16     Mat (){ memset(a,0,sizeof(a)); }
17 }a,res;
18 
19 Mat mul(Mat a,Mat b){
20     Mat c;
21     rep(i,0,ed) rep(j,0,ed) rep(k,0,ed) c[i][k]+=a[i][j]*b[j][k];
22     return c;
23 }
24 
25 Mat ksm(Mat a,int b){
26     Mat res;
27     rep(i,0,ed) res[i][i]=1;
28     for (; b; a=mul(a,a),b>>=1)
29         if (b & 1) res=mul(res,a);
30     return res;
31 }
32 
33 int main(){
34     freopen("bzoj4000.in","r",stdin);
35     freopen("bzoj4000.out","w",stdout);
36     scanf("%d%d%d%d",&n,&m,&p,&k); k++; ed=(1<<m)-1;
37     rep(i,1,3) rep(j,1,p){
38         scanf("%d",&x);
39         if (x) d[++tot]=(P){i-2,j-k};
40     }
41     rep(S1,0,ed) rep(S2,0,ed){
42         bool flag=0;
43         rep(i,0,m-1) if (S1&(1<<i))
44             rep(j,0,m-1) if (S2&(1<<j))
45                 rep(k,1,tot) if ((d[k].x==1 && d[k].y==j-i) || (d[k].x==-1 && d[k].y==i-j))
46                     { flag=1; break; }
47         rep(i,0,m-2) if (S2&(1<<i))
48             rep(j,i+1,m-1) if (S2&(1<<j))
49                 rep(k,1,tot) if (d[k].x==0 && (d[k].y==j-i || d[k].y==i-j))
50                     { flag=1; break; }
51         a[S1][S2]=!flag;
52     }
53     res[0][0]=1; res=mul(res,ksm(a,n));
54     rep(i,0,ed) ans+=res[0][i];
55     printf("%u\n",ans);
56     return 0;
57 }

 

转载于:https://www.cnblogs.com/HocRiser/p/9870643.html