D. Big Brush

Posted 2022-02-22 thusloop

Big Brush
题意:每次涂色2×2的矩阵 （可以覆盖），给出目标矩阵，求涂色方案。
思路：反过来想，最后一次涂色一定 是2×2的矩阵内颜色相同（可以多个起点），反过来bfs
修改已经处理过的矩阵为0；

//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int inf=8e18;
const int maxn=2e5+100;
int a[1100][1100],n,m;
bool vis[1100][1100];
int dir[4][2]= 0,0,1,0,0,1,1,1;
int dir2[8][2]= -1,-1  ,-1,0  ,-1,1  ,0,-1  ,1,-1  ,0,1  ,1,1  ,1,0;
void out()

	cout<<"\\n";
	for(int i=1;i<=n;i++)
	
		for(int j=1;j<=m;j++)
		
			cout<<a[i][j]<<" ";
		
		cout<<"\\n";
	
	cout<<"\\n";

int ck(int x,int y)

	//bool fg=1;
	if(x<=0||x>=n||y<=0||y>=m)return 0;
	int num=0;
	for(int i=0; i<4; i++)
	
		int x1=x+dir[i][0];
		int y1=y+dir[i][1];
		if(x1<=0||x1>n||y1<=0||y1>m)continue;
		if(a[x1][y1]!=0)num=a[x1][y1];
	
	for(int i=0; i<4; i++)
	
		int x1=x+dir[i][0];
		int y1=y+dir[i][1];
		if(x1<=0||x1>n||y1<=0||y1>m)continue;
		if(a[x1][y1]!=0&&a[x1][y1]!=num)return 0;
	
	return num;

void gui(int x,int y)

	for(int i=0; i<4; i++)
	
		int x1=x+dir[i][0];
		int y1=y+dir[i][1];
		if(x1<=0||x1>n||y1<=0||y1>m)continue;
		a[x1][y1]=0;
	

int x,y;
stack<tuple<int,int,int>>s;
void sol(int x,int y)

	queue<pii>q;
	q.push(x,y);
	vis[x][y]=1;
	//s.push(x,y,ck(x,y));
	while(!q.empty())
	
		pii now=q.front();
		q.pop();
		int num=ck(now.fi,now.se);
		vis[now.fi][now.se]=0;
		if(num==0)continue;
		s.push(now.fi,now.se,num);
		gui(now.fi,now.se);
		for(int i=0; i<8; i++)
		
			int x1=now.fi+dir2[i][0];
			int y1=now.se+dir2[i][1];
			if(x1<=0||x1>=n||y1<=0||y1>=m)continue;
			if(vis[x1][y1]==0)q.push(x1,y1);vis[x1][y1]=1;
		
		//out();
	

signed main()

	IOS
	cin>>n>>m;
	for(int i=1; i<=n; i++)
		for(int j=1; j<=m; j++)
			cin>>a[i][j];
	vector<pii>v;		
	for(int i=1; i<n; i++)
	
		for(int j=1; j<m; j++)
		
			if(ck(i,j))
			
				x=i;
				y=j;
				v.pb(i,j);
			
		
	
	if(x==0&&y==0)
	
		cout<<"-1\\n";
		return 0;
	
	for(auto it:v) sol(it.fi,it.se);
	//sol();
	//bool fg=1;
	for(int i=1; i<=n; i++)
	
		for(int j=1; j<=m; j++)
		
			if(a[i][j]!=0)
			
				cout<<"-1\\n";
				return 0;
			
		
	
	cout<<s.size()<<"\\n";
	while(!s.empty())
	
		auto now=s.top();
		s.pop();
		cout<<get<0>(now)<<" "<<get<1>(now)<<" "<<get<2>(now)<<"\\n";