$[TJOI2017]$ 可乐 矩阵优化$dp$

\(Sol\)

设\(f_i\)为到第\(i\)秒的方案数,显然\(f_i=\)在第\(i\)秒前爆炸的方案数+在第\(i\)秒爆炸的方案数+在第\(i\)秒停下的方案数+在第\(i\)秒走向下一个城市

的方案数.注意到第四个转移和当前在哪个城市有关,所以要另记一维\(j\)表示当前位置.于是\(f_{i,j}=\)第\(i\)秒前在\(j\)爆炸的方案数+第\(i\)秒在\(j\)爆炸的方案数+第\(i\)秒停在\(j\)的方案数+第\(i\)秒由别的城市走向\(j\)的方案数.记这四个量分别为\(f1,f2,f3,f4\).

\(f1_{i,j}=f1_{i-1,j}+f2_{i-1,j}\)

\(f2_{i,j}=f3_{i-1,j}+f4_{i-1,j}\)

\(f3_{i,j}=f3_{i-1,j}+f4_{i-1,j}\)

\(f4_{i,j}=f3_{i-1,k}+f4_{i-1,k}\),其中,\(k\)是\(j\)的相邻城市.

这样瞎\(dp\)一下就可以获得\(20pts\)的好成绩\(QAQ\).

其实上面的转移方程看起来就很矩阵优化的亚子,于是矩阵优化一下就好辣.

\(Code\)

#include<bits/stdc++.h>
#define il inline
#define Ri register int
#define go(i,a,b) for(Ri i=a;i<=b;++i)
#define yes(i,a,b) for(Ri i=a;i>=b;--i)
#define e(i,u) for(Ri i=b[u];i;i=a[i].nt)
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define db double
#define inf 2147483647
using namespace std;
il int read()
{
    Ri x=0,y=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*y;
}
const int mod=2017;
int n,m,t,as;
vector<int>to[31];
struct mt
{
    int a[121][121];bool ste;
    il void clear(){mem(a,0);ste=0;}
}trs,cur;
il void inc(Ri &x,Ri y){x+=y;if(x>=mod)x-=mod;}
il mt operator * (mt x,mt y)
{
    mt z;z.clear(),z.ste=x.ste;Ri h=x.ste?1:n*4;
    go(i,1,h)
	go(j,1,n*4)
	go(k,1,n*4)
	inc(z.a[i][j],1ll*x.a[i][k]*y.a[k][j]%mod);
    return z;
}
il void init()
{
    trs.clear();
    go(i,1,n)
    {
	    Ri mi=(i-1)*4+1;
	    trs.a[mi][mi]=trs.a[mi+1][mi]=1;
	    trs.a[mi+2][mi+1]=trs.a[mi+3][mi+1]=1;
	    trs.a[mi+2][mi+2]=trs.a[mi+3][mi+2]=1;
    	    go(j,0,(int)to[i].size()-1){Ri k=(to[i][j]-1)*4+3;trs.a[k][mi+3]=trs.a[k+1][mi+3]=1;}
    }
}
int main()
{
    n=read(),m=read();
    go(i,1,m){Ri u=read(),v=read();to[u].push_back(v),to[v].push_back(u);}
    init();cur.clear();cur.a[1][3]=1,cur.ste=1;
    t=read();
    while(t){if(t&1)cur=cur*trs;trs=trs*trs;t>>=1;}
    go(i,1,n*4)inc(as,cur.a[1][i]);
    printf("%d\n",as);
    return 0;
}