SSL2512&BZOJ1898 沼泽鳄鱼

想明白了这东西能分成12个时刻，于是就有12个矩阵，然后乘起来当成操作就行了，时间复杂度一个n3logk

Problem Id:2511 User Id:BPM136 

Memory:1756K  Time:75MS
Language:G++  Result:Accepted

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define LL long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
inline LL read()
{
	LL d=0,f=1;char s=getchar();
	while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
	while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
	return d*f;
}
#define N 50
#define FISH 25
#define inf 10000
int n,m,nm;

struct matrix
{
	int a[N][N];
	void clear()
	{
		memset(a,0,sizeof(a));
	}
	void OUT()
	{
		fo(i,0,m-1)
		{
			fo(j,0,m-1)
			{
				cout<<a[i][j]<<' ';
			}cout<<endl;
		}cout<<endl;
	}
	matrix operator*(const matrix b)const
	{
		matrix anss;
		fo(i,0,m-1)
		fo(j,0,m-1)
		{
			anss.a[i][j]=0;
			fo(k,0,m-1)
			anss.a[i][j]+=a[i][k]*b.a[k][j];
			anss.a[i][j]%=inf;
		}
		return anss;
	}
}I,tot[FISH+5],A;
int fish[FISH][5],s,e,nfish;

void getI()
{
	fo(i,0,m-1)
	fo(j,0,m-1)
	if(i==j)I.a[i][j]=1;
	else I.a[i][j]=0;
}

matrix KSM(matrix a,int k)
{
	matrix ret=I;
	while(k)
	{
		if(k&1)ret=a*ret;
		a=a*a;
		k>>=1;
	}
	return ret;
}

int main()
{
	m=read(),n=read(),s=read(),e=read();nm=read();
	fo(i,1,12)tot[i].clear();
	fo(i,1,n)
	{
		int x=read(),y=read();
		fo(j,1,12)
		tot[j].a[x][y]=tot[j].a[y][x]=1;
	}
	nfish=read();getI();A=I;
	fo(i,1,nfish)
	{
		fish[i][0]=read();
		fo(j,1,fish[i][0])fish[i][j]=read();
		fo(j,1,12)
		{
			fo(k,0,m-1)
			tot[j].a[k][fish[i][j%fish[i][0]+1]]=0;
		}
	}
//	fo(i,1,12)tot[i].OUT();
	fo(i,1,12)A=A*tot[i];
	A=KSM(A,nm/12);nm%=12;
	fo(i,1,nm)A=A*tot[i];
	cout<<A.a[s][e]<<endl;
	return 0;
}