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; }
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