D. Big Brush

// Problem: D. Big Brush
// Contest: Codeforces - Codeforces Round #771 (Div. 2)
// URL: https://codeforces.com/contest/1638/problem/D
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 2022-02-15 20:19:54
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}

const int N = 1e3 + 10;

int n, m;
int a[N][N];
int ans[N * N][3];
bool used[N][N];
int cc[N];
int cnt;
void check (int x, int y) {
	if (x < 0 || x + 1 >= n || y < 0 || y + 1 >= m) return;
	if (used[x][y]) return;
	
	int sz = 0;
	for (int i = 0; i < 2; i ++) {
		for (int j = 0; j < 2; j ++) {
			int c = a[x + i][y + j];
			if (c != -1) {
				cc[sz++] = c;
			}
		}
	}
	if (sz == 0) return;
	sort(cc, cc + sz);
	if (cc[0] != cc[sz -  1]) return;
	ans[cnt][0] = x;
	ans[cnt][1] = y;
	ans[cnt][2] = cc[0];
	used[x][y] = 1;
	cnt++;
}
void solve() {
	scanf("%d%d", & n, & m);
	for (int i = 0 ; i< n; i ++) {
		for (int j = 0; j < m; j ++)
			scanf("%d", & a[i][j]);
	}
	for (int i = 0; i < n - 1; i++)
		for (int j  =0; j < m - 1; j++)
			check(i, j);
		
	for (int i  = 0; i < cnt;i  ++) {
		int x = ans[i][0], y = ans[i][1];
		a[x][y] = a[x + 1][y] = a[x + 1][ y + 1] = a[x][y + 1] = -1;
		for (int j = -1; j <= 1; j++) {
			for (int k = -1; k <= 1; k ++) {
				check(x + j, k + y);
			}	
		}
	}
	bool ok = 1;
	for (int i = 0; i < n; i ++)
		for (int j = 0; j < m ;j ++)
			ok &= a[i][j] == -1;
	
	if (!ok) {
		puts("-1");
		return;
	}
	printf("%d\n",cnt);
	for (int i = cnt - 1; i >= 0; i --)
		printf("%d %d %d\n", ans[i][0] + 1, ans[i][1] + 1, ans[i][2]);
		
}
int main () {
	solve();
    // int t;
    // t = 1;
    // while (t --) solve();
    return 0;
}

By enumerating the last operation , The last operation must be four squares of the same color . And then you can push back through this square , By querying the square containing any one of the squares . Then judge whether the colors inside are the same , If it's all the same , Then it can be used as the second step , Push class by this . This question is mainly about reverse thinking Why not ？ Because we didn't find that after removing the current square , Just judge whether the colors in other squares are the same , There is no need to judge every stain . That's too much trouble . In the future, we should understand the nature