// 沿着矩阵的中间垂直线翻转矩阵，使其变为自身的镜像。
void mirror(VS& s)
{
    for (int i = 0; i < n; i ++ )
        for (int j = 0, k = n - 1; j < k; j ++ , k -- )
            swap(s[i][j], s[i][k]);
}

// 将矩阵顺时针旋转 90 度。
void rotate(VS& s)
{
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < i; j ++ )
            swap(s[i][j], s[j][i]);
    mirror(s);
}

矩阵

#include <iostream>
#include <cstring>
using namespace std;
template <int row, int col>
struct Matrix
{
    int r, c;
    long long ele[row][col];
    Matrix() { throw "Please set row and column!"; }
    Matrix(int a, int b) : r(a), c(b) { memset(ele, 0, sizeof(ele)); }
    long long &operator()(int a, int b) { return ele[a][b]; }
    Matrix &operator+=(Matrix oth) { return *this = *this + oth; }
    Matrix &operator-=(Matrix oth) { return *this = *this - oth; }
    Matrix &operator*=(Matrix oth) { return *this = *this * oth; }
};
template <int m, int n>
Matrix<m, n> operator+(Matrix<m, n> m1, Matrix<m, n> m2)
{
    Matrix<m, n> ret;
    for (int i = 0; i < m1.r; i++)
        for (int j = 0; j < m1.c; j++)
            ret(i, j) = m1(i, j) + m2(i, j);
    return ret;
}
template <int m, int n>
Matrix<m, n> operator-(Matrix<m, n> m1, Matrix<m, n> m2)
{
    Matrix<m, n> ret;
    for (int i = 0; i < m1.r; i++)
        for (int j = 0; j < m1.c; j++)
            ret(i, j) = m1(i, j) + m2(i, j);
    return ret;
}
template <int m, int n, int p>
Matrix<m, p> operator*(Matrix<m, n> m1, Matrix<n, p> m2) //普通乘法
{
    Matrix<m, p> ret(m1.r, m2.c);
    for (int i = 0; i < m1.r; i++)
        for (int k = 0; k < m1.c; k++)
            for (int j = 0; j < m2.c; j++)
                ret(i, j) += m1(i, k) * m2(k, j);
    return ret;
}
template <int m, int n>
Matrix<m, n> operator^(Matrix<m, n> mat, long long k) //快速幂
{
    Matrix<m, n> ans = mat;
    for (k--; k; mat *= mat, k >>= 1)
        if (k & 1)
            ans *= mat;
    return ans;
}
template <int m, int n, int p>
Matrix<m, p> mulmod(Matrix<m, n> m1, Matrix<n, p> m2, int mod) //乘加取模
{
    Matrix<m, p> ret(m1.r, m2.c);
    for (int i = 0; i < m1.r; i++)
        for (int k = 0; k < m1.c; k++)
            for (int j = 0; j < m2.c; j++)
                ret(i, j) = (ret(i, j) + m1(i, k) * m2(k, j) % mod) % mod;
    return ret;
}
template <int m, int n>
Matrix<m, n> qpowmod(Matrix<m, n> mat, long long k, int mod) //快速幂取模
{
    Matrix<m, n> ans = mat;
    for (k--; k; mat = mulmod(mat, mat, mod), k >>= 1)
        if (k & 1)
            ans = mulmod(ans, mat, mod);
    return ans;
}
template <int m, int n>
ostream &operator<<(ostream &os, Matrix<m, n> &mat)
{
    for (int i = 0; i < mat.r; i++)
    {
        for (int j = 0; j < mat.c; j++)
            os << mat(i, j) << ' ';
        os << endl;
    }
    return os;
}
template <int m, int n>
istream &operator>>(istream &is, Matrix<m, n> &mat)
{
    for (int i = 0; i < mat.r; i++)
        for (int j = 0; j < mat.c; j++)
            is >> mat(i, j);
    return is;
}
const int maxn = 105;
int main()
{
    int n;
    long long k;
    const int p = 1e9 + 7;
    cin >> n >> k;
    Matrix<maxn, maxn> mat(n, n);
    cin >> mat;
    auto tmp = qpowmod(mat, k, p);
    cout << tmp;
    return 0;
}

最大牛（流）

EKKKK
bool bfs()
{
    int hh = 0, tt = 0;
    memset(st, false, sizeof st);
    q[0] = S, st[S] = true, d[S] = INF;
    while (hh <= tt)
    {
        int t = q[hh ++ ];
        for (int i = h[t]; ~i; i = ne[i])
        {
            int ver = e[i];
            if (!st[ver] && f[i])
            {
                st[ver] = true;
                d[ver] = min(d[t], f[i]);
                pre[ver] = i;
                if (ver == T) return true;
                q[ ++ tt] = ver;
            }
        }
    }
    return false;
}

int EK()
{
    int r = 0;
    while (bfs())
    {
        r += d[T];
        for (int i = T; i != S; i = e[pre[i] ^ 1])
            f[pre[i]] -= d[T], f[pre[i] ^ 1] += d[T];
    }
    return r;
}

bool bfs()
{
    int hh = 0, tt = 0;
    memset(d, -1, sizeof d);
    q[0] = S, d[S] = 0, cur[S] = h[S];
    while (hh <= tt)
    {
        int t = q[hh ++ ];
        for (int i = h[t]; ~i; i = ne[i])
        {
            int ver = e[i];
            if (d[ver] == -1 && f[i])
            {
                d[ver] = d[t] + 1;
                cur[ver] = h[ver];
                if (ver == T)  return true;
                q[ ++ tt] = ver;
            }
        }
    }
    return false;
}

int find(int u, int limit)
{
    if (u == T) return limit;
    int flow = 0;
    for (int i = cur[u]; ~i && flow < limit; i = ne[i])
    {
        cur[u] = i;  // 当前弧优化
        int ver = e[i];
        if (d[ver] == d[u] + 1 && f[i])
        {
            int t = find(ver, min(f[i], limit - flow));
            if (!t) d[ver] = -1;
            f[i] -= t, f[i ^ 1] += t, flow += t;
        }
    }
    return flow;
}

int dinic()
{
    int r = 0, flow;
    while (bfs()) while (flow = find(S, INF)) r += flow;
    return r;
}

2175. 飞行员配对方案问题 求方案
    for (int i = 0; i < idx; i += 2)
        if (e[i] > m && e[i] <= n && !f[i])
            printf("%d %d\n", e[i ^ 1], e[i]);

    // 用于保存节点状态的数组
    // 递归遍历子节点
    // 如果是父节点则跳过
    // 递归遍历子节点
    // 根据实际问题更新节点状态
    // 根据实际问题更新节点状态
    // 构建树形结构
    // 初始化节点状态
    // 从根节点开始遍历
    // 输出结果
作者：北海没有WA
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来源：AcWing
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