# One car comes and one car goes

听说是小学奥数题，没啥好说的

# Color

裸的树形 DP ，穷举一下每个点选各种 type 的种类数就好，剩下的就是乘法原理吧

b.cpp

	#include <bits/stdc++.h>

	#define MOD 10000009
	#define N 50000
	#define M 30

	using namespace std;

	int n, m, s, t;

	bool can[N][M];
	vector<int> a[N];
	long long f[N][M];

	inline long long nis(int c, int i) {
	static int ret;
	ret = f[c][0] - f[c][i];
	return ret < 0 ? ret + MOD : ret;
	}

	void dfs(int t, int p) {
	int len = a[t].size(), c;
	for (int i = 0; i < len; i++)
	if (a[t][i] != p) {
	dfs(a[t][i], t);
	}
	for (int i = 1; i <= m; i++) {
	if (can[t][i])
	f[t][i] = 1;
	else continue ;
	for (int j = 0; j < len; j++) {
	if ((c = a[t][j]) == p)
	continue ;
	f[t][i] *= nis(c, i);
	f[t][i] %= MOD;
	}
	}
	for (int i = 1; i <= m; i++) {
	f[t][0] += f[t][i];
	f[t][0] %= MOD;
	}
	}

	int main() {
	srand(time(0));
	ios::sync_with_stdio(0);

	while (cin >> n >> m) {
	memset(f, 0, sizeof(f));
	for (int i = 1; i <= n; i++)
	a[i].clear();
	for (int i = 1; i < n; i++) {
	cin >> s >> t;
	a[s].push_back(t);
	a[t].push_back(s);
	}
	for (int i = 1; i <= n; i++)
	for (int j = 1; j <= m; j++) {
	cin >> t;
	can[i][j] = t;
	}

	dfs(s = rand() % n + 1, 0);
	cout << f[s][0] << endl;
	}

	return 0;
	}

给一个无向图，问在 T 步之内有多少条路线从 1 走到 n

结论题，直接建立图的邻接矩阵，然后求矩阵 T 次幂就好

由于题目要求到了 n 就不能走出去，所以需要删掉 n 所有出边，然后 n 自环一下就好

但是由于太弱并不知道还有自环这种操作，一股脑删光了所有射出边，给邻接矩阵多加了一层，用一种十分不优雅的方式侥幸过了 XD

e.cpp

	#include <bits/stdc++.h>

	#define N 100
	#define MOD 1000000007LL

	using namespace std;

	typedef long long LL;
	typedef vector<vector<LL>> mat;

	int n, m;

	inline void init(mat &x, bool I) {
	x.resize(n + 1);
	for (int i = 0; i <= n; i++)
	x[i].resize(n + 1);
	for (int i = 0; i <= n; i++)
	for (int j = 0; j <= n; j++)
	x[i][j] = 0;
	if (I) for (int i = 0; i <= n; i++)
	x[i][i] = 1;
	}

	inline mat mul(const mat &a, const mat &b) {
	mat c; init(c, false);
	for (int i = 0; i <= n; i++)
	for (int j = 0; j <= n; j++) {
	for (int k = 0; k <= n; k++) {
	c[i][j] += (a[i][k] * b[k][j]) % MOD;
	c[i][j] %= MOD;
	}
	}
	return c;
	}

	inline mat quick_mod(mat a, int q) {
	mat c; init(c, true);
	while (q) {
	if (q & 1)
	c = mul(a, c);
	a = mul(a, a);
	q >>= 1;
	}
	return c;
	}

	inline void print(const mat &a) {
	for (int i = 0; i <= n; i++, cout << endl)
	for (int j = 0; j <= n; j++)
	cout << a[i][j] << " ";
	}

	int main() {
	int x, y, t;
	cin >> n >> m;
	mat a; init(a, false);
	while (m--) {
	cin >> x >> y;
	x--, y--;
	a[x][y] = 1;
	a[y][x] = 1;
	}
	cin >> t;

	for (int i = 0; i < n; i++)
	a[n - 1][i] = 0;
	a[n - 1][n] = 1;
	a[n][n] = 1;

	a = quick_mod(a, t);
	a[0][n] += a[0][n - 1];
	cout << a[0][n] % MOD << endl;

	return 0;
	}