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

Posted 2021-01-20 acgoto

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：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=105;
 4 const int inf=500;
 5 int n,m,cnt1,cnt2,ans,sum,mp[maxn][maxn],min_row[maxn],min_col[maxn];
 6 pair<int,int> paii_row[maxn],paii_col[maxn];
 7 int main(){
 8     while(cin>>n>>m){
 9         ans=sum=cnt1=cnt2=0;
10         for(int i=1;i<=n;++i)min_row[i]=inf;
11         for(int j=1;j<=m;++j)min_col[j]=inf;
12         for(int i=1;i<=n;++i){
13             for(int j=1;j<=m;++j){
14                 cin>>mp[i][j],sum+=mp[i][j];
15                 min_row[i]=min(min_row[i],mp[i][j]);
16                 min_col[j]=min(min_col[j],mp[i][j]);
17             }
18         }
19         if(n>m){///如果行大于列，则从列开始操作
20             for(int j=1;j<=m;++j){
21                 for(int i=1;i<=n;++i){
22                     mp[i][j]-=min_col[j];
23                     min_row[i]=min(min_row[i],mp[i][j]);
24                 }
25                 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;
26             }
27             for(int i=1;i<=n;++i){///从行开始操作
28                 for(int j=1;j<=m;++j){
29                     mp[i][j]-=min_row[i];
30                     min_col[j]=min(min_col[j],mp[i][j]);
31                 }
32                 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;
33             }
34         }
35         else{
36             for(int i=1;i<=n;++i){///否则先从行开始操作
37                 for(int j=1;j<=m;++j){
38                     mp[i][j]-=min_row[i];
39                     min_col[j]=min(min_col[j],mp[i][j]);
40                 }
41                 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;
42             }
43             for(int j=1;j<=m;++j){
44                 for(int i=1;i<=n;++i){
45                     mp[i][j]-=min_col[j];
46                     min_row[i]=min(min_row[i],mp[i][j]);
47                 }
48                 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;
49             }
50         }
51         if(sum)cout<<-1<<endl;
52         else{
53             cout<<ans<<endl;
54             for(int i=0;i<cnt1;++i)
55                 for(int j=1;j<=paii_row[i].first;++j)
56                     cout<<"row "<<paii_row[i].second<<endl;
57             for(int i=0;i<cnt2;++i)
58                 for(int j=1;j<=paii_col[i].first;++j)
59                     cout<<"col "<<paii_col[i].second<<endl;
60         }
61     }
62     return 0;
63 }

 

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