HUAS 1482 lsy的后宫(DP+矩阵快速幂)

这道题的DP是很好想的，令dp[i][j]表示第i个位置摆第j种妹子的方法数，j为0表示不摆妹子的方法数。

dp[i][j]=sigma(dp[i-1][k])(s[j][k]!='1').容易看出这是个递推式，于是可以用矩阵快速幂加速DP转移。

复杂度O(m^3*logn).

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

struct Matrix{LL matrix[N][N];}a, sa, unit, kk;
char s[N][N];
int n, m;

Matrix Mul(Matrix a, Matrix b) //矩阵乘法(%MOD)
{
    Matrix c;
    FOR(i,,m) FOR(j,,m) {
          c.matrix[i][j]=;
          FOR(l,,m) c.matrix[i][j]+=(a.matrix[i][l]*b.matrix[l][j])%MOD;
          c.matrix[i][j]%=MOD;
    }
    return c;
}
Matrix Cal(int exp)  //矩阵快速幂
{
    Matrix p=a, q=unit;
    if (exp==) return q;
    while (exp!=) {
        if (exp&) exp--, q=Mul(p,q);
        else exp>>=, p=Mul(p,p);
    }
    return Mul(p,q);
}
void init(){
    FOR(i,,m) unit.matrix[i][i]=kk.matrix[][i]=;
    FOR(i,,m) a.matrix[i][]=a.matrix[][i]=;
    FOR(i,,m) FOR(j,,m) if (s[i][j]=='') a.matrix[j][i]=;
}
int main ()
{
    LL ans=;
    scanf("%d%d",&n,&m);
    FOR(i,,m) scanf("%s",s[i]+);
    init();
    sa=Cal(n-); sa=Mul(kk,sa);
    FOR(i,,m) ans=(ans+sa.matrix[][i])%MOD;
    printf("%lld\n",ans);
    return ;
}