思路:不能走走过来的路,变点交换跑矩阵快速幂。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; const int N = 1e5 + 7; const int M = 5e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 45989; int n, m, t, S, T, U[N], V[N], b[N]; struct Matrix { int a[125][125], n; Matrix(int _n) { n = _n; memset(a, 0, sizeof(a)); } void init() { for(int i = 0; i < n; i++) a[i][i] = 1; } Matrix operator * (const Matrix &B) const { Matrix C(n); for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) for(int k = 0; k < n; k++) C.a[i][j] = (C.a[i][j] + a[i][k] * B.a[k][j]) % mod; return C; } Matrix operator ^ (int b) { Matrix A = (*this); Matrix ans(n); ans.init(); while(b) { if(b & 1) ans = ans * A; A = A * A; b >>= 1; } return ans; } }; int main() { scanf("%d%d%d%d%d", &n, &m, &t, &S, &T); for(int i = 1; i <= m; i++) { scanf("%d%d", &U[i << 1], &V[i << 1]); U[i << 1 | 1] = V[i << 1]; V[i << 1 | 1] = U[i << 1]; } m = m << 1 | 1; Matrix A(m + 1); for(int i = 2; i <= m; i++) { if(U[i] == S) A.a[0][i] = 1; if(V[i] == T) A.a[i][1] = 1; for(int j = 2; j <= m; j++) if(V[i] == U[j] && (i ^ j) != 1) A.a[i][j] = 1; } A = A ^ (t + 1); printf("%d ", A.a[0][1]); return 0; }