【题解】 bzoj1875: [SDOI2009]HH去散步 （动态规划+矩阵乘法）

//It is coded by Ning_Mew on 5.10
#include<bits/stdc++.h>
#define IL inline
using namespace std;

const int maxn=125,MOD=45989;

int n,m,t,A,B,out=0;
struct Blc{
  int a[maxn][maxn];
  Blc(){memset(a,0,sizeof(a));}
}bas;
int head[maxn],cnt=0;
struct Edge{
  int nxt,to;
}edge[maxn];

void add(int from,int to){
  edge[++cnt].nxt=head[from];edge[cnt].to=to;head[from]=cnt;
}
IL int o(int x){if(x%2)return x+1;return x-1;}
IL Blc X(Blc x,Blc y){
  Blc ans;
  for(int i=1;i<=2*m;i++){
    for(int j=1;j<=2*m;j++){
      for(int k=1;k<=2*m;k++){
        ans.a[i][j]+=x.a[i][k]*y.a[k][j]%MOD;
        ans.a[i][j]%=MOD;
      }
    }
  }return ans;
}
IL Blc q_pow(Blc x,int s){
  Blc ans=bas;
  while(s){
    if(s%2)ans=X(ans,x);
    x=X(x,x);
    s=s/2;
  }return ans;
}
int main(){
  scanf("%d%d%d%d%d",&n,&m,&t,&A,&B);
  for(int i=1;i<=m;i++){
    int x,y;scanf("%d%d",&x,&y);
    add(x,y);add(y,x);
  }
  Blc box;
  for(int u=0;u<n;u++){
    for(int i=head[u];i!=0;i=edge[i].nxt){
      int v=edge[i].to;
      for(int ii=head[v];ii!=0;ii=edge[ii].nxt){
    //cout<<"pr:"<<i<<‘ ‘<<ii<<endl;
    if(i==o(ii))continue;
    bas.a[i][ii]++;
      }
    }
  }
  //pr(bas);
  for(int i=head[A];i!=0;i=edge[i].nxt){box.a[1][i]++;}  
  
  Blc ans=q_pow(bas,t-2);
  
  ans=X(box,ans);

  for(int i=1;i<=2*m;i++){
    if(edge[i].to==B){out=(out+ans.a[1][i])%MOD;}
  }
  //cout<<endl;pr(ans);
  printf("%d\n",out);
  return 0;
}