【ZOJ3256】Tour in the Castle （插头dp）

Tour in the Castle

题意

After the final BOSS is defeated, the hero found that the whole castle is collapsing (very familiar scene, isn't it). Escape from the castle is easy, just need to cross a few rooms. But as the Hero is full of adventurous spirit, he decides to visit every room before he escape the castle.

The castle is a rectangle with N * M rooms in it. Two rooms are connected if they share a common edge. The hero starts in the top left room. And the bottom left room is the only way out. After the hero visits a room and leaves it, it will collapse immediately(Another familiar scene). So he can visit each room only once.

The diagram shows one tour over a castle with 4 * 10 rooms:

思路

看m的范围，可以用矩阵加速

原来是逐格递推，现在要改成逐行递推。

考虑合法状态，因为平常记录的是括号序列，所以枚举状态的时候判断是否是合法的括号序列并直接枚举算之后的转移状态。

转移的时候要注意如果此时没有插头一定是无法继续向下转移的，或者上面已经形成回路后也不能向下转移。

//ZOJ 3256
#include<bits/stdc++.h>
using namespace std;
const int Mod=7777777;
int id=0,a[8],pos[17005],Id[135],n,m,b[8],inv[8];
void Put(int x) {
	if(x>=10)Put(x/10);
	putchar(x%10+'0');
}
void Print(int x) {
	Put(x),puts("");
}
int get(int *A) {
	int now=0;
	for(int i=n; i>=1; i--)now=now*4+A[i];
	return now;
}
void Printf(int x) {
	for(int i=1; i<=n; i++)printf("%d",x%4),x>>=2;
	puts("");
}
void add(int &x,int y) {
	x+=y;
	if(x>=Mod)x-=Mod;
	if(x<0)x+=Mod;
}
struct node {
	int A[135][135];
	void clear() {
		memset(A,0,sizeof(A));
	}
	void Init() {
		for(int i=0; i<id; i++)for(int j=0; j<id; j++)A[i][j]=(i==j);
	}
	node operator*(const node&_)const {
		node res;
		res.clear();
		for(int i=0; i<id; i++)
			for(int j=0; j<id; j++)
				for(int k=0; k<id; k++)
					add(res.A[i][k],1ll*A[i][j]*_.A[j][k]%Mod);
		return res;
	}
} T,B;
node mul(node A,int y) {
	node res;
	res.Init();
	while(y) {
		if(y&1)res=res*A;
		A=A*A,y>>=1;
	}
	return res;
}
void dfs(int x,int la) {
	if(x==n+1) {
		if(la)return;
		pos[get(a)]=id,Id[id++]=get(a);
		return;
	}
	if(la)a[x]=2,dfs(x+1,la-1);
	a[x]=1,dfs(x+1,la+1),a[x]=0,dfs(x+1,la);
}
void redfs(int x,int now,int rt) {
	if(x==n+1) {
		add(T.A[rt][pos[now]],1);
		return;
	}
	int b1=(now>>(2*(x-1)))%4,b2=(now>>(2*x))%4;
	if(!(b1|b2)) {
		if(x!=n)redfs(x+1,now+inv[x-1]+2*inv[x],rt);
	} else if(b1&&!b2) {
		if(x!=n)redfs(x+1,now-b1*inv[x-1]+b1*inv[x],rt);
		redfs(x+1,now,rt);
	} else if(b2&&!b1) {
		if(x!=n)redfs(x+1,now,rt);
		redfs(x+1,now+b2*inv[x-1]-b2*inv[x],rt);
	} else if(b1==1&&b2==1) {
		int k1=1;
		for(int i=x+1; i<=n; i++) {
			if(((now>>(2*i))%4)==1)k1++;
			else if(((now>>(2*i))%4)==2)k1--;
			if(!k1) {
				redfs(x+1,now-inv[x-1]-inv[x]-inv[i],rt);
				break;
			}
		}
	} else if(b1==2&&b2==2) {
		int k1=1;
		for(int i=x-2; i>=0; i--) {
			if(((now>>(2*i))%4)==1)k1--;
			else if(((now>>(2*i))%4)==2)k1++;
			if(!k1) {
				redfs(x+1,now-inv[x-1]*2-inv[x]*2+inv[i],rt);
				break;
			}
		}
	} else if(b1==2&&b2==1)redfs(x+1,now-b1*inv[x-1]-b2*inv[x],rt);
	else if((x==n)&&(now-inv[x-1]*b1-inv[x]*b2==0))redfs(x+1,now-inv[x-1]*b1-inv[x]*b2,rt);
}
void Init() {
	id=0,dfs(1,0);
	int res=(1<<(2*n+2))-1;
	for(int i=0; i<id; i++) {
		if(!Id[i])continue;
		redfs(1,(Id[i]<<2)&res,i);
	}
}
int main() {
	inv[0]=1;
	for(int i=1; i<=7; i++)inv[i]=inv[i-1]<<2;
	while(~scanf("%d%d",&n,&m)) {
		T.clear(),Init();
		T=mul(T,m);
		int p=inv[0]+2*inv[n-1];
		p=pos[p],B.clear(),B.A[0][p]=1,B=B*T;
		if(B.A[0][pos[0]]==0)puts("Impossible");
		else printf("%d\n",B.A[0][pos[0]]);
	}
	return 0;
}