【jzoj4870】【涂色游戏】【动态规划】【矩阵快速幂】

题目大意

解题思路

设f[i][j]表示一列有i个数，j种颜色的方案数，f[i][j]=f[i-1][j-1]*(p-j+1)+f[i-1][j]*j。g[i][j]表示第i列，j种颜色的方案数，g[i][j]=g[i-1][k]*mat[j][k]。设x表示i，j并集，mat[i][j]=f[n][j]/c[p][j]*c[i][i+j-x]*c[p-i][x-i]。推出矩阵后就可以快速幂。

code

#include<set>#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long long#define LD double#define max(a,b) ((a>b)?a:b)#define min(a,b) ((a>b)?b:a)#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)using namespace std;int const inf=1e9;int const maxn=100;int n,m,p,q;LL f[maxn+10][maxn+10],ni[maxn+10],c[maxn+10][maxn+10],ans[maxn+10][maxn+10],mat[maxn+10][maxn+10],tmp[maxn+10][maxn+10],mod=998244353;void multansmat(){    fo(i,1,maxn)fo(j,1,maxn)tmp[i][j]=0;    fo(i,1,maxn)fo(j,1,maxn)fo(k,1,maxn)        tmp[i][k]=(tmp[i][k]+ans[i][j]*mat[j][k])%mod;    fo(i,1,maxn)fo(j,1,maxn)ans[i][j]=tmp[i][j];}void multmatmat(){    fo(i,1,maxn)fo(j,1,maxn)tmp[i][j]=0;    fo(i,1,maxn)fo(j,1,maxn)fo(k,1,maxn)        tmp[i][k]=(tmp[i][k]+mat[i][j]*mat[j][k])%mod;    fo(i,1,maxn)fo(j,1,maxn)mat[i][j]=tmp[i][j];}LL Pow(LL x,LL y){    LL z=1;    for(;y;){        if(y&1)z=(z*x)%mod;        x=(x*x)%mod;        y/=2;    }    return z;}int main(){    //freopen("color.in","r",stdin);    //freopen("color.out","w",stdout);    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d%d%d%d",&n,&m,&p,&q);    fo(i,0,maxn){        c[i][0]=1;        fo(j,1,i)            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;    }    fo(i,1,p)ni[i]=Pow(c[p][i],mod-2);    f[0][0]=1;    fo(i,1,n)        fo(j,1,min(i,p))            f[i][j]=(f[i-1][j-1]*(p-j+1)%mod+f[i-1][j]*j%mod)%mod;    fo(i,1,p)        fo(j,1,p){            fo(x,max(q,max(i,j)),min(p,i+j))                mat[i][j]=(mat[i][j]+f[n][j]*ni[j]%mod*c[i][i+j-x]%mod*c[p-i][x-i]%mod)%mod;            //mat[i][j]=(mat[i][j])%mod;        }    fo(i,1,p)ans[1][i]=f[n][i];m--;    for(;m;){        if(m&1)multansmat();        multmatmat();        m/=2;    }    LL anss=0;    fo(i,1,p)anss=(anss+ans[1][i])%mod;    printf("%lld",anss);    return 0;}