CodeForces - 816C Karen and Game（简单模拟）

Problem Description

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!

Input

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).

Output

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.

Examples

input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1

output

4
row 1
row 1
col 4
row 3

input

3 3
0 0 0
0 1 0
0 0 0

output

-1

input

3 3
1 1 1
1 1 1
1 1 1

output

3
row 1
row 2
row 3

Note

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.

解题思路：贪心模拟，行操作完再进行列操作。注意：题目要求用最少的步数，因此当n>m时，应先对列进行贪心减操作，再对行进行操作；否则先对行进行贪心减操作，最后如果减不完，则直接输出-1.

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
const int inf=500;
int n,m,cnt1,cnt2,ans,sum,mp[maxn][maxn],min_row[maxn],min_col[maxn];
pair<int,int> paii_row[maxn],paii_col[maxn];
int main(){
    while(cin>>n>>m){
        ans=sum=cnt1=cnt2=0;
        for(int i=1;i<=n;++i)min_row[i]=inf;
        for(int j=1;j<=m;++j)min_col[j]=inf;
        for(int i=1;i<=n;++i){
            for(int j=1;j<=m;++j){
                cin>>mp[i][j],sum+=mp[i][j];
                min_row[i]=min(min_row[i],mp[i][j]);
                min_col[j]=min(min_col[j],mp[i][j]);
            }
        }
        if(n>m){///如果行大于列，则从列开始操作
            for(int j=1;j<=m;++j){
                for(int i=1;i<=n;++i){
                    mp[i][j]-=min_col[j];
                    min_row[i]=min(min_row[i],mp[i][j]);
                }
                if(min_col[j])sum-=min_col[j]*n,ans+=(paii_col[cnt2].first=min_col[j]),paii_col[cnt2++].second=j,min_col[j]=0;
            }
            for(int i=1;i<=n;++i){///从行开始操作
                for(int j=1;j<=m;++j){
                    mp[i][j]-=min_row[i];
                    min_col[j]=min(min_col[j],mp[i][j]);
                }
                if(min_row[i])sum-=min_row[i]*m,ans+=(paii_row[cnt1].first=min_row[i]),paii_row[cnt1++].second=i,min_row[i]=0;
            }
        }
        else{
            for(int i=1;i<=n;++i){///否则先从行开始操作
                for(int j=1;j<=m;++j){
                    mp[i][j]-=min_row[i];
                    min_col[j]=min(min_col[j],mp[i][j]);
                }
                if(min_row[i])sum-=min_row[i]*m,ans+=(paii_row[cnt1].first=min_row[i]),paii_row[cnt1++].second=i,min_row[i]=0;
            }
            for(int j=1;j<=m;++j){
                for(int i=1;i<=n;++i){
                    mp[i][j]-=min_col[j];
                    min_row[i]=min(min_row[i],mp[i][j]);
                }
                if(min_col[j])sum-=min_col[j]*n,ans+=(paii_col[cnt2].first=min_col[j]),paii_col[cnt2++].second=j,min_col[j]=0;
            }
        }
        if(sum)cout<<-1<<endl;
        else{
            cout<<ans<<endl;
            for(int i=0;i<cnt1;++i)
                for(int j=1;j<=paii_row[i].first;++j)
                    cout<<"row "<<paii_row[i].second<<endl;
            for(int i=0;i<cnt2;++i)
                for(int j=1;j<=paii_col[i].first;++j)
                    cout<<"col "<<paii_col[i].second<<endl;
        }
    }
    return 0;
}