On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!
The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).
If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.
The next k lines should each contain one of the following, describing the moves in the order they must be done:
- row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row".
- col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column".
If there are multiple optimal solutions, output any one of them.
3 5 2 2 2 3 2 0 0 0 1 0 1 1 1 2 1
4 row 1 row 1 col 4 row 3
3 3 0 0 0 0 1 0 0 0 0
-1
3 3 1 1 1 1 1 1 1 1 1
3 row 1 row 2 row 3
In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.
#include<bits/stdc++.h>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
const int N=2e5+5;
const int MOD=1e9+9;
const int INF=0x3f3f3f3f;
int a[200][200],n,m;
vector <pair<string,int> > ans;
bool checkrow(int i,int j){
for(int jj=1;jj<=m;jj++)
if(a[i][jj]==0) return false;
return true;
}
bool checkcol(int i,int j){
for(int ii=1;ii<=n;ii++)
if(a[ii][j]==0) return false;
return true;
}
int main(void){
cin >>n >>m ;
for(int i=1;i<=n;++i)
for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i][j]==0) continue;
while(a[i][j]){
bool row=checkrow(i,j);
bool col=checkcol(i,j);
if(row && col){
if(n>m){
for(int ii=1;ii<=n;ii++)
a[ii][j]--;
ans.push_back(make_pair("col",j));
}
else{
for(int jj=1;jj<=m;jj++)
a[i][jj]--;
ans.push_back(make_pair("row",i));
}
}
else if(row && !col){
for(int jj=1;jj<=m;jj++)
a[i][jj]--;
ans.push_back(make_pair("row",i));
}
else if(!row && col){
for(int ii=1;ii<=n;ii++)
a[ii][j]--;
ans.push_back(make_pair("col",j));
}
else if(!row && !col) break;
}
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i][j]!=0){
cout << -1 << endl;
return 0;
}
}
}
cout <<(int) ans.size()<<endl;
for(int i=0;i<(int)ans.size();++i){
cout << ans[i].first <<" "<<ans[i].second <<endl;
}
return 0;
}