可以发现，如果没有鳄鱼，那么就是裸地一道题，但是可以发现鳄鱼最多每12次重复，那么就少于12的那部分dp，其他的就矩阵乘法就行了
PS：第一次吧矩阵乘法AC了好开心QAQ
CODE：
#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#define mod 10000#define maxn 60using namespace std;
struct mat{
int n,m,a[maxn][maxn];
mat(){n=m=0;memset(a,0,sizeof(a));}
int I(int _x){n=m=_x;for (int i=1;i<=n;i++) a[i][i]=1;}
};mat operator *(const mat &x,const mat &y){
mat ans;
ans.n=x.n;
ans.m=y.m;
for (int i=1;i<=ans.n;i++)
for (int j=1;j<=ans.m;j++)
for (int k=1;k<=x.m;k++){
ans.a[i][j]+=x.a[i][k]*y.a[k][j];
ans.a[i][j]%=mod;
}
return ans;
}mat power(mat x,int y){
mat ans;ans.I(x.n);
for (;y;y>>=1){
if (y&1) ans=ans*x;
x=x*x;
}
return ans;
}int n,m,s,e,fn,ti;
bool b[60][60];
int a[20][7],t[60],f[60][60][15];
int main(){
scanf("%d%d%d%d%d",&n,&m,&s,&e,&ti);
for (int i=1;i<=m;i++) {
int x,y;
scanf("%d%d\n",&x,&y);
x++;y++;
b[x][y]=b[y][x]=1;
}
scanf("%d",&fn);
for (int i=1;i<=fn;i++) {
scanf("%d",&t[i]);
for (int j=1;j<=t[i];j++) {scanf("%d",&a[i][j]);a[i][j]++; }
}
for (int i=1;i<=n;i++) f[i][i][0]=1;
for (int i=1;i<=12;i++)
for (int j=1;j<=n;j++)
for (int k=1;k<=n;k++){
for (int l=1;l<=n;l++)
if (f[j][l][i-1]&&b[l][k]) (f[j][k][i]+=f[j][l][i-1])%=mod;
for (int l=1;l<=fn;l++)
if (a[l][i%t[l]+1]==k) {f[j][k][i]=0;}
}
mat x,y;
x.n=x.m=y.n=y.m=n;
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
x.a[i][j]=f[i][j][12];
x=power(x,ti/12);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
y.a[i][j]=f[i][j][ti%12];
x=x*y;
printf("%d",x.a[s+1][e+1]);
return 0;
} |