bzoj 1838

让邻接矩阵做乘法，对于不能走的，因为t只有234，所以构造12个矩阵然后分k/12和k%12做就行

#include<bits/stdc++.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define inf 10000
#define ll long long
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define NM 100
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
struct mat{
    int a[NM][NM];
}a[13],ans,null;
int n,m,S,T,k,p,_x,_y,_t;
mat operator*(const mat&x,const mat&y){
    mat s;
    mem(s.a);
    inc(i,1,n)
    inc(j,1,n)
    inc(k,1,n)(s.a[i][j]+=x.a[i][k]*y.a[k][j])%=inf;
    return s;
}
mat mul(int k){
    mat s=null,t=a[0];
    for(;k;k>>=1,t=t*t)if(k&1)s=s*t;
    return s;
}
void out(mat t){
    inc(i,1,n){
        inc(j,1,n)printf("%d ",t.a[i][j]);putchar('
');
    }putchar('
');
}
int main(){
    freopen("data.in","r",stdin);
    n=read();m=read();S=read()+1;T=read()+1;k=read();
    mem(null.a);
    inc(i,1,n)null.a[i][i]=1;
    inc(i,1,m){
        _x=read()+1;_y=read()+1;
        a[1].a[_x][_y]++;a[1].a[_y][_x]++;
    }
    inc(i,2,12)a[i]=a[1];
    p=read();
    while(p--){
        _t=read();
        inc(i,0,_t-1){
            _x=read()+1;
            for(int j=i;j<=12;j+=_t)if(j)
            inc(k,1,n)a[j].a[k][_x]=0;
        }
    }
    a[0]=a[1];inc(i,2,12)a[0]=a[0]*a[i];
    ans=mul(k/12);
    inc(i,1,k%12)ans=ans*a[i];
    printf("%d
",ans.a[S][T]);
    return 0;
}