Bzoj 1875 [sdoi] hh go for a walk (DP, matrix multiplication optimization)

Title Link: Bzoj 1875

The idea of this problem, the main idea is to construct a matrix is very ingenious. Our common use of matrix multiplication transfer is to use points to transfer, but this does not remove the situation in a place to stay.
A very magical approach is to use the edge to compose (for an edge I, except for an edge J to meet i== (j^1), and other edges are connected) transfer t-1 times,
It is then multiplied by a matrix (equivalent to a coefficient matrix) constructed from an edge that is connected to the starting point. Finally, just add the answer to the side that connects the end point.

#include <cstdio> #include <cstring> #include <iostream> using namespace std;
#define MOD (45989) int n,m,t,a,b,tot=0,k=2;

int q[10000],head[10000]; struct Edge{int v,next;}

E[125];
	struct node{int a[125][125];
	Node () {memset (a,0,sizeof (a));

}}s;
	inline int read () {int X=0,f=1;char ch=getchar (); while (ch< ' 0 ' | |
	Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}
	while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();}
return x*f;
	} node Mut (node X,node y) {node t; for (int i=0;i<k;i++) for (int j=0;j<k;j++) for (int p=0;p<k;p++) t.a[i][j]= (t.a[i][j]+x.a[i][p] * y.a[p][j
	])%mod;
return t;
	} Node Pow (node X,int y) {node t;
	for (int i=0;i<=k;i++) t.a[i][i]=1;
		while (y) {if (y&1) T=mut (t,x);
		y>>=1;
	X=mut (X,X);
} return t;

} void Adde (int u,int v) {e[k].v=v; e[k].next=head[u]; head[u]=k++;} void input () {n=read (); M=read (); T=read (); A=read ();
	B=read ();
		for (int i=1;i<=m;i++) {int x=read (), Y=read (); Adde (x, y); ADDE (Y,X);
	
	}} void Solve () {for (int i=head[a];i;i=e[i].next) s.a[0][i]=1;
	Node T;
		for (int i=2;i<k;i++) {int v=e[i].v;
		if (v==b) q[++tot]=i;
			for (int j=head[v];j;j=e[j].next) {if (i== (j^1)) continue;
		T.a[i][j]=1;
	}} node Sai=pow (t,t-1);
	Node Ans=mut (S,SAI);
	int cnt=0;
	for (int i=1;i<=tot;i++) {cnt= (Cnt+ans.a[0][q[i]])%mod;
} printf ("%d\n", CNT);
	} int main () {input ();
	Solve ();
return 0; }