[SDOI2009] HH去散步
设 \(dp_{i_j}\) 表示第 \(i\) 个时刻，走到第 \(j\) 条边的终点的方案数
转移：从当前点 \(u\) 开始，枚举从 \(u\) 出发的所有点，向 \(dp_{i+1,to_u}\) 转移
但是 \(t\) 非常大，所以需要使用矩阵快速幂
设初始矩阵为 \(\text{A}\)，转移矩阵为 \(\text{B}\)，那么答案矩阵 \(\text{C} = \text{A}\times\text{B}^{t-1}\)
// Problem: P2151 [SDOI2009] HH去散步
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2151
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// https://codeforces.com/problemset/customtest# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
# define int long long
# define rd(t) read <t> ()
# define mem(a, b) memset (a, b, sizeof (a))
# define fi first
# define se second
# define lc u << 1
# define rc u << 1 | 1
# define debug printf ("debug\n")
const int N = 155, M = N, mod = 45989;
template <typename T> inline T read ()
{T s = 0; int w = 1; char c = getchar (); for (; !isdigit (c); c = getchar ()) { if (c == '-') w = -1; }for (; isdigit (c); c = getchar ()) s = (s << 1) + (s << 3) + c - '0';return s * w;
}int n, m, t, A, B;
struct edge { int vtx, nxt; } e[M]; int h[N], idx;
void add (int u, int v) { e[ ++ idx] = {v, h[u]}; h[u] = idx; }
struct matrix
{int a[N][N]; matrix () { mem (a, 0); } matrix operator * (const matrix &b) const{matrix ans; for (int i = 1; i < N; i ++ ){for (int k = 1; k < N; k ++ ){for (int j = 1; j < N; j ++ )ans.a[i][j] = (ans.a[i][j] + a[i][k] * b.a[k][j]) % mod; }}return ans; }
} a, b;
matrix quick_pow (matrix ans, matrix a, int b)
{while (b){if (b & 1) ans = ans * a; a = a * a; b >>= 1; }return ans;
}
int rev (int x) { return (x % 2 == 0 ? x - 1 : x + 1); }
signed main ()
{mem (h, -1); n = rd (int), m = rd (int), t = rd (int); A = rd (int) + 1, B = rd (int) + 1; for (int i = 1; i <= m; i ++ ){int u = rd (int) + 1, v = rd (int) + 1; add (u, v), add (v, u); } for (int i = 1; i <= idx; i ++ ){int u = e[i].vtx; for (int j = h[u]; ~j; j = e[j].nxt){if (i == rev (j)) continue; b.a[i][j] ++ ; }}for (int i = h[A]; ~i; i = e[i].nxt) a.a[1][i] ++ ; matrix ans = quick_pow (a, b, t - 1); int res = 0; for (int i = h[B]; ~i; i = e[i].nxt)res = (res + ans.a[1][rev (i)]) % mod; printf ("%lld\n", res); return 0;
}
