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.

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.

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题目的意思是给出一个全0矩阵，操作是在一行或一列全+1，问怎样才能得到给定矩阵

思路：从给定矩阵出发，每一行每一列能全减就贪心减去，根据行列数量决定先处理行还是列

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <cmath>

using namespace std;

#define LL long long

const int inf=0x7fffffff;

int mp[500][500];

struct node

{

    int f,x,nu;

} ans[200000];

int n,m,cnt,tot;

void Row()

{

    for(int i=0; i<m; i++)

    {

        int mn=inf;

        for(int j=0; j<n; j++)

        {

            mn=min(mn,mp[i][j]);

        }

        for(int j=0; j<n; j++)

        {

            mp[i][j]-=mn;

        }

        if(mn>0)

        {

            tot+=mn;

            ans[cnt].f=1;

        ans[cnt].x=mn;

        ans[cnt++].nu=i+1;

        }

    }

}

void Col()

{

    for(int j=0; j<n; j++)

    {

        int mn=inf;

        for(int i=0; i<m; i++)

        {

            mn=min(mn,mp[i][j]);

        }

        for(int i=0; i<m; i++)

        {

            mp[i][j]-=mn;

        }

        if(mn>0)

        {

            tot+=mn;

            ans[cnt].f=2;

        ans[cnt].x=mn;

        ans[cnt++].nu=j+1;

        }

    }

}

int main()

{

    scanf("%d%d",&m,&n);

    for(int i=0; i<m; i++)

        for(int j=0; j<n; j++)

            scanf("%d",&mp[i][j]);

    cnt=0;

    tot=0;

    if(m>n)

    {

        Col();

        Row();

    }

    else

    {

        Row();

        Col();

    }

int flag=0;

    for(int i=0; i<m; i++)

        for(int j=0; j<n; j++)

        {

            if(mp[i][j]>0)

            {

                flag=1;

                break;

            }

            if(flag)

                break;

        }

    if(flag)

        printf("-1\n");

    else

    {

        printf("%d\n",tot);

        for(int i=0; i<cnt; i++)

        {

            if(ans[i].f==1)

                for(int j=0;j<ans[i].x;j++)

                printf("row %d\n",ans[i].nu);

            else

            for(int j=0;j<ans[i].x;j++)

                printf("col %d\n",ans[i].nu);

        }

    }

    return 0;

}