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）

 1 #include<bits/stdc++.h>
 2 #define inc(i,l,r) for(int i=l;i<=r;i++)
 3 #define dec(i,l,r) for(int i=l;i>=r;i--)
 4 #define link(x) for(edge *j=h[x];j;j=j->next)
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define inf 45989
 7 #define ll long long
 8 #define succ(x) (1<<x)
 9 #define NM 300
10 using namespace std;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(isdigit(ch))x=x*10+ch-‘0‘,ch=getchar();
15     return x*f;
16 }
17 struct mat{
18     int a[NM][NM];
19 }c,_t;
20 int n,m,S,T,a[NM],b[NM],ans;
21 mat operator*(const mat&x,const mat&y){
22     mat s;mem(s.a);
23     inc(i,1,m)
24     inc(j,1,m)
25     inc(k,1,m)
26     (s.a[i][j]+=x.a[i][k]*y.a[k][j]%inf)%=inf;
27     return s;
28 }
29 ll k;
30 int main(){
31     freopen("data.in","r",stdin);
32     n=read();m=read();k=read();S=read()+1;T=read()+1;
33     inc(i,1,m){
34         a[i*2-1]=b[i*2]=read()+1;b[i*2-1]=a[i*2]=read()+1;
35     }
36     m<<=1;
37     inc(i,1,m)c.a[i][i]=1;
38     inc(i,1,m)
39     inc(j,1,m)
40     if((i-1)/2!=(j-1)/2&&b[i]==a[j])_t.a[i][j]=1;
41     /*inc(i,1,m){
42     inc(j,1,m)printf("%d ",_t.a[i][j]);printf("\n");}*/
43     for(k--;k;k>>=1,_t=_t*_t)
44     if(k&1)c=c*_t;
45 /*    inc(i,1,m){
46     inc(j,1,m)printf("%d ",c.a[i][j]);printf("\n");}*/
47     inc(i,1,m)if(a[i]==S)
48     inc(j,1,m)if(b[j]==T)
49     (ans+=c.a[i][j]%inf)%=inf;
50     printf("%d\n",ans);
51     return 0;
52 }