一眼看成了 [NOIP2020] 移球游戏，不过这题比移球游戏简单多了。
考虑如何在不破坏已归位煎饼的同时将一个新的煎饼归位。
首先找到第一个未归位的位置，以及一个应该放到这个位置的煎饼，例如下面这样：（每一列为一叠煎饼，上面是上面，下面是下面，字符 . 表示空）
1 a . 1 b .
[x] (1) . 1 c . 1 d .
-----------
可以通过以下七步将 (1) 放到第一列中：
- 将
(1)上面的煎饼移动到空白列；
1 . . 1 . .
[x] (1) . 1 c a 1 d b
-----------
- 将
(1)移动到空白列；
1 . . 1 . .
[x] . (1)1 c a 1 d b
-----------
- 将空白列移动到第二列（以上三步的目的是将
(1)移动到所在列顶端）；
1 (1) . 1 a .
[x] b . 1 c . 1 d .
-----------
- 将
[x]及以上的煎饼移动到空白列；
. (1) . . a . . b 1 1 c 1 1 d [x]
-----------
- 将
(1)移动到第一列；
. . . . a .
(1) b 1 1 c 1 1 d [x]
-----------
- 将
[x]上面的煎饼移动到第一列；
1 . . 1 a .
(1) b . 1 c . 1 d [x]
-----------
- 将
[x]移动到第二列。
1 [x] . 1 a .
(1) b . 1 c . 1 d .
-----------
至此，我们把 (1) 归位了。
我们至多需要归位 \(nm\) 个煎饼，因此操作次数最多为 \(7nm\)。
代码：
// Problem: P15346 [TOIP 2025] 煎饼摊
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P15346
// Memory Limit: 512 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {uniform_int_distribution<int> dist(L, R);return dist(rnd);
}template<typename T> void chkmin(T& x, T y) {if(y < x) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}template<int mod>
inline unsigned int down(unsigned int x) {return x >= mod ? x - mod : x;
}template<int mod>
struct Modint {unsigned int x;Modint() = default;Modint(unsigned int x) : x(x) {}friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}friend Modint operator/(Modint a, Modint b) {return a * ~b;}friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}friend Modint operator~(Modint a) {return a ^ (mod - 2);}friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}friend Modint& operator++(Modint& a) {return a += 1;}friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}friend Modint& operator--(Modint& a) {return a -= 1;}friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}friend bool operator==(Modint a, Modint b) {return a.x == b.x;}friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};const int N = 55;int n, m, a[N][N], sz[N];
vector<tuple<int, int, int>> ans;void operate(int x, int y, int z) {if(y > sz[x]) return;ans.emplace_back(x, sz[x] - y + 1, z);rep(i, y, sz[x]) a[z][++sz[z]] = a[x][i];sz[x] = y - 1;
}int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);cin >> m >> n;rep(i, 1, n) rep(j, 1, m) cin >> a[i][j];rep(i, 1, n) sz[i] = m;while(true) {int ax = -1, ay = -1;rep(i, 1, n) {rep(j, 1, m) {if(a[i][j] != i) {ax = i;ay = j;break;}}if(ax != -1 && ay != -1) break;}if(ax == -1 && ay == -1) break;int bx = -1, by = -1;rep(i, 1, n) {rep(j, 1, m) {if(i != ax && a[i][j] == ax) {bx = i;by = j;break;}}if(bx != -1 && by != -1) break;}operate(bx, by + 1, n + 1);operate(bx, by, n + 1);operate(n + 1, 1, bx);operate(ax, ay, n + 1);operate(bx, m, ax);operate(n + 1, 2, ax);operate(n + 1, 1, bx);}cout << ans.size() << endl;for(auto [x, y, z]: ans) cout << x << " " << y << " " << z << endl;// rep(i, 1, n + 1) {// cout << "Stack #" << i << ": ";// rep(j, 1, sz[i]) cout << a[i][j] << " ";// cout << endl;// }return 0;
}