HDU 5411(CRB and Puzzle-矩阵A+A^2+..+A^n)

CRB and Puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 678 Accepted Submission(s): 253



[align=left]Problem Description[/align]
CRB is now playing Jigsaw Puzzle.

There are N
kinds of pieces with infinite supply.

He can assemble one piece to the right side of the previously assembled one.

For each kind of pieces, only restricted kinds can be assembled with.

How many different patterns he can assemble with at most
M
pieces? (Two patterns P
and Q
are considered different if their lengths are different or there exists an integer
j
such that j-th
piece of P
is different from corresponding piece of Q.)

[align=left]Input[/align]
There are multiple test cases. The first line of input contains an integer
T,
indicating the number of test cases. For each test case:

The first line contains two integers N,
M
denoting the number of kinds of pieces and the maximum number of moves.

Then N
lines follow. i-th
line is described as following format.

k a1 a2 ... ak

Here k
is the number of kinds which can be assembled to the right of the
i-th
kind. Next k
integers represent each of them.

1 ≤ T
≤ 20

1 ≤ N
≤ 50

1 ≤ M
≤ 105

0 ≤ k
≤ N

1 ≤ a1
< a2
< … < ak
≤ N

[align=left]Output[/align]
For each test case, output a single integer - number of different patterns modulo 2015.

[align=left]Sample Input[/align]

1
3 2
1 2
1 3
0

[align=left]Sample Output[/align]

6
Hintpossible patterns are ∅, 1, 2, 3, 1→2, 2→3

[align=left]Author[/align]
KUT（DPRK）

[align=left]Source[/align]
2015 Multi-University Training Contest 10

[align=left]Recommend[/align]
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显然矩阵DP，记得A+A^2+..+A^N 可以用矩阵快速二分优化

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (2015)
#define eps (1e-3)
#define MAXN (50+10)
typedef __int64 ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

struct M
{
int n,m;
ll a[MAXN][MAXN];
M(int _n=0){n=m=_n;MEM(a);}
M(int _n,int _m){n=_n,m=_m;MEM(a);}
void mem (int _n=0){n=m=_n;MEM(a);}
void mem (int _n,int _m){n=_n,m=_m;MEM(a);}

friend M operator*(M a,M b)
{
M c;
c.mem(a.n,b.m) ;
For(k,a.m)
For(i,a.n)
For(j,b.m)
c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%F;
return c;
}
friend M operator+(M a,M b)
{
For(i,a.n)
For(j,a.m)
a.a[i][j]=(a.a[i][j]+b.a[i][j])%F;
return a;
}

void make_I(int _n)
{
n=m=_n; MEM(a)
For(i,n) a[i][i]=1;
}
}f;

M a;
int n,m;

bool a2[1000000];
M pow2(M a,ll b)
{
M c;c.make_I(a.n);
int n=0;while (b) a2[++n]=b&1,b>>=1;
For(i,n)
{
if (a2[i]) c=c*a;
a=a*a;
}
return c;
}

bool a3[1000000];
M pow222(M a,ll b)
{
M c;c.make_I(a.n);
int n=0;while (b) a3[++n]=b&1,b>>=1;
c=a; b=1;
M d=c;
ForD(i,n-1)
{
b=b*2+a3[i];
c=c*d+c;
d=d*d;
if (a3[i]) c=c*a+a,d=d*a;
}
return c;
}

M pow22(M a,ll b)
{
M c;c.make_I(a.n);
if (b==0) return c;
if (b==1) return a;
c=pow22(a,b/2);
c=c*pow2(a,b/2)+c;
if (b&1) c=c*a+a;

return c;
}

int main()
{
//	freopen("Puzzle.in","r",stdin);

int T; cin>>T;
while(T--) {
scanf("%d%d",&n,&m);
f.mem(n);
For(i,n) {
int k;scanf("%d",&k);
For(j,k)
{
int p;
scanf("%d",&p);
f.a[p][i]=1;
}
}
if (m==1)
{
cout<<n+1<<endl;
continue;
}

//		f.pri();
f=pow222(f,m-1);
//		f.pri();

ll ans=1+n;
For(i,n) For(j,n) upd(ans,f.a[i][j]);

printf("%I64d\n",ans);
}

return 0;
}