題面

題目傳送門

解法

如果沒有不能經過上一次經過的邊這個限制，顯然就是矩陣乘法的裸題

那麽我們考慮轉化一下，把邊當成點

將一條無向邊拆成2條有向邊，然後連邊，設鄰接矩陣為\(A\)

將\(A\)變成\(A^{T-1}\)，然後枚舉起點的出邊，終點的入邊即可

時間復雜度：\(O(m^3\ log\ T)\)

代碼

#include <bits/stdc++.h>
#define Mod 45989
#define N 150
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
    x = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == ‘-‘) f = -1; c = getchar();}
    while (isdigit(c)) x = x * 10 + c - ‘0‘, c = getchar(); x *= f;
}
struct Node {
    int x, y;
} a[N];
struct Matrix {
    int a[N][N];
    void Clear() {memset(a, 0, sizeof(a));}
};
int n, m, T, s, t, tot, cnt;
vector <int> e[N];
Matrix operator * (Matrix x, Matrix y) {
    Matrix ret; ret.Clear();
    for (int k = 2; k <= tot; k++)
        for (int i = 2; i <= tot; i++)
            for (int j = 2; j <= tot; j++)
                ret.a[i][j] = (ret.a[i][j] + x.a[i][k] * y.a[k][j] % Mod) % Mod;
    return ret;
}
Matrix operator ^ (Matrix x, int y) {
    Matrix ret = x; y--;
    while (y) {
        if (y & 1) ret = ret * x;
        y >>= 1, x = x * x;
    }
    return ret;
}
int main() {
    read(n), read(m), read(T), read(s), read(t);
    s++, t++, tot = 1;
    for (int i = 1; i <= m; i++) {
        int x, y; read(x), read(y);
        a[++tot] = (Node) {++x, ++y};
        a[++tot] = (Node) {y, x};
        e[x].push_back(tot - 1), e[x].push_back(tot);
        e[y].push_back(tot - 1), e[y].push_back(tot);
    }
    Matrix tx; tx.Clear();
    for (int i = 1; i <= n; i++)
        for (int j = 0; j < e[i].size(); j++)
            for (int k = 0; k < e[i].size(); k++) {
                int x = e[i][j], y = e[i][k];
                if (x == y || (x ^ 1) == y) continue;
                if (a[x].y == a[y].x) tx.a[x][y] = 1;
            }
    tx = tx ^ (T - 1); int ans = 0;
    for (int i = 2; i <= tot; i++)
        for (int j = 2; j <= tot; j++)
            if (a[i].x == s && a[j].y == t) ans = (ans + tx.a[i][j]) % Mod;
    cout << ans << "\n";
    return 0;
}
 

bzoj 1875 [SDOI2009]HH去散步 矩陣乘法