BZOJ 1875: [SDOI2009]HH去散步

直接对点集进行矩阵快速幂不好解决不能走回头路的限制
那么考虑对边集进行矩阵快速幂
即把边看成点，枚举相邻的边进行连边
再矩阵快速幂即可

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define lowbit(i) ((i) & (-i))
#define ll long long
#define ull unsigned long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=0,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 45989;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 132, E = 132;
Edg
struct Mat {
	int mat[E][E];
	Mat() { memset(mat, 0, sizeof(mat)); }
	Mat operator * (const Mat &p) const {
		Mat res;
		rep (i, 0, ccnt + 1) {
			rep (j, 0, ccnt + 1) {
				ll temp = 0;
				rep (k, 0, ccnt + 1) {
					temp += (ll)mat[i][k] * p.mat[k][j];
					if (temp >= (1LL << 60)) temp %= MOD;
				}
				res.mat[i][j] = temp % MOD;
			}
		}
		return res;
	}
};
Mat qp(Mat a, int b) {
	Mat c;
	rep (i, 0, ccnt + 1) c.mat[i][i] = 1;
	for (; b; a = a * a, b >>= 1) if (b & 1) c = c * a;
	return c;
}
int n, m, k, s, t;
inline int rev (int x) {
	if (x & 1) x++;
	else x--;
	return x;
}

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	n = _(), m = _(), k = _(), s = _(), t = _();
	rep (i, 0, m) {
		int a = _(), b = _();
		add(a, b);
	}
	Mat base;
	es (s, i, v) base.mat[0][i]++;
	Mat res;
	rep (i, 1, ccnt + 1) {
		int v = to[i];
		es (v, j, u) {
			if (j != rev(i)) res.mat[i][j] = 1;
		}
	}
	res = qp(res, k - 1);
	base = base * res;
	int ans = 0;
	rep (i, 1, ccnt + 1) if (to[i] == t) M(ans += base.mat[0][i]);
	printf("%d\n", ans);
	return 0;
}