bzoj 1875 [SDOI2009]HH去散步 矩阵乘法

题面

题目传送门

解法

如果没有不能经过上一次经过的边这个限制，显然就是矩阵乘法的裸题

那么我们考虑转化一下，把边当成点

将一条无向边拆成2条有向边，然后连边，设邻接矩阵为\(A\)

将\(A\)变成\(A^{T-1}\)，然后枚举起点的出边，终点的入边即可

时间复杂度：\(O(m^3\ log\ T)\)

代码

#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;
}