bzoj 1875

把边当做一个状态（orz居然有这种想法），初始点也看成边，然后矩阵快速幂就可以了（感觉有点像flyod）

#include<bits/stdc++.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define inf 45989
#define ll long long
#define succ(x) (1<<x)
#define NM 300
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
struct mat{
    int a[NM][NM];
}c,_t;
int n,m,S,T,a[NM],b[NM],ans;
mat operator*(const mat&x,const mat&y){
    mat s;mem(s.a);
    inc(i,1,m)
    inc(j,1,m)
    inc(k,1,m)
    (s.a[i][j]+=x.a[i][k]*y.a[k][j]%inf)%=inf;
    return s;
}
ll k;
int main(){
    freopen("data.in","r",stdin);
    n=read();m=read();k=read();S=read()+1;T=read()+1;
    inc(i,1,m){
        a[i*2-1]=b[i*2]=read()+1;b[i*2-1]=a[i*2]=read()+1;
    }
    m<<=1;
    inc(i,1,m)c.a[i][i]=1;
    inc(i,1,m)
    inc(j,1,m)
    if((i-1)/2!=(j-1)/2&&b[i]==a[j])_t.a[i][j]=1;
    /*inc(i,1,m){
    inc(j,1,m)printf("%d ",_t.a[i][j]);printf("
");}*/
    for(k--;k;k>>=1,_t=_t*_t)
    if(k&1)c=c*_t;
/*    inc(i,1,m){
    inc(j,1,m)printf("%d ",c.a[i][j]);printf("
");}*/
    inc(i,1,m)if(a[i]==S)
    inc(j,1,m)if(b[j]==T)
    (ans+=c.a[i][j]%inf)%=inf;
    printf("%d
",ans);
    return 0;
}