CodeForces 816C 思维

On the way to school, Karen became fixated on the puzzle game on her phone!

$\text{[math]}$

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 g_i, 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 g_i, j (0 ≤ g_i, 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:

rowx, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row".
colx, (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.

Sample Input

Input

Output

4
row 1
row 1
col 4
row 3

Input

Output

-1

Input

Output

3
row 1
row 2
row 3

Hint

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:

$\text{[math]}$

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:

$\text{[math]}$

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

　　分析这种棋盘的特性，要加加一行，要加加一列，且不能减，且棋盘初始都为零。所以如果同一行上出现了不同的数，则数大的一列必然有该列的加，行同理。

所以可以先处理列，把所有列加过的都还原回去，所以只剩下了由个别的行改变所导致的棋盘，再用同样的方法将行还原回去，最后如果棋盘仍然不为零的话，说明所有的行都进行过相同次数的

加（这里要注意！！！如果最后棋盘不为零的话，既有可能是所有的行进行了操作，也有可能是所有的列进行了操作，因为题目让求最少的操作次数，所以应判断行和列谁小操作的谁）。

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<string>
  5 #include<cmath>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<stack>
  9 #include<deque>
 10 #include<map>
 11 #include<iostream>
 12 using namespace std;
 13 typedef long long  LL;
 14 const double pi=acos(-1.0);
 15 const double e=exp(1);
 16 const int N = 100009;
 17 
 18 LL con[110][110];
 19 LL check[110][110];
 20 LL col[110],row[110];
 21 
 22 int main()
 23 {
 24     LL i,p,j,m,n;
 25     LL flag=0;
 26     scanf("%lld%lld",&m,&n);
 27     LL head=10000000,tail=10000000,mid;
 28 
 29     for(i=1; i<=m; i++)
 30     {
 31         for(j=1; j<=n; j++)
 32         {
 33             scanf("%lld",&con[i][j]);
 34             if(con[1][j]<head)
 35                 head=con[1][j];
 36             if(con[i][1]<tail)
 37                 tail=con[i][1];
 38         }
 39     }
 40 
 41     if(m<=n)
 42     {
 43         for(i=1; i<=n; i++)
 44         {
 45             if(con[1][i]>head)
 46                 col[i]+=con[1][i]-head;
 47         }
 48         mid=tail-col[1];
 49         for(i=1; i<=m; i++)
 50         {
 51             if(con[i][1]>tail)
 52                 row[i]+=con[i][1]-tail;
 53             row[i]+=mid;
 54         }
 55     }
 56     else
 57     {
 58         for(i=1; i<=m; i++)
 59         {
 60             if(con[i][1]>tail)
 61                 row[i]+=con[i][1]-tail;
 62         }
 63         mid=head-row[1];
 64         for(i=1; i<=n; i++)
 65         {
 66             if(con[1][i]>head)
 67                 col[i]+=con[1][i]-head;
 68             col[i]+=mid;
 69         }
 70     }
 71 
 72     flag=0;
 73 
 74     for(i=1; i<=n; i++)
 75     {
 76         if(col[i])
 77         {
 78             flag+=col[i];
 79             for(j=1; j<=m; j++)
 80                 check[j][i]+=col[i];
 81         }
 82     }
 83     for(i=1; i<=m; i++)
 84     {
 85         if(row[i])
 86         {
 87             flag+=row[i];
 88             for(j=1; j<=n; j++)
 89                 check[i][j]+=row[i];
 90         }
 91     }
 92 
 93     for(i=1; i<=m; i++)
 94     {
 95         for(j=1; j<=n; j++)
 96         {
 97             if(con[i][j]!=check[i][j])
 98                 break;
 99         }
100         if(j<=n)
101         {
102             flag=-1;
103             break;
104         }
105     }
106 
107 
108 
109     if(flag==-1)
110         printf("-1\n");
111     else
112     {
113         printf("%lld\n",flag);
114         for(i=1; i<=n; i++)
115         {
116             while(col[i])
117             {
118                 col[i]--;
119                 printf("col %lld\n",i);
120             }
121         }
122         for(i=1; i<=m; i++)
123         {
124             while(row[i])
125             {
126                 row[i]--;
127                 printf("row %lld\n",i);
128             }
129         }
130     }
131     return 0;
132 }

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