Codeforces Round #419 (Div. 2) C. Karen and Game

时间:2024-09-29 17:07:44
C. Karen and Game
time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

Codeforces Round #419 (Div. 2) C. Karen and Game

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:

Codeforces Round #419 (Div. 2) C. Karen and Game

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:

Codeforces Round #419 (Div. 2) C. Karen and Game

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

 题意:

给定n行m列数字,每次可以让一行或一列都减一,求出让全部数字全为0的最小的次数,没有则输出-1;

思路:

就是对行和列处理,注意下n和m的大小就行。

第一次比赛打了三道。。。上了一波大分。

#include <bits/stdc++.h>

using namespace std;
const int N = ; int n, m, a[N][N], c[N], r[N], ans; int check() {
for (int i = ; i < n; i++)
for (int j = ; j < m; j++)
if (a[i][j]) return ;
return ;
} void row_man() {
for (int i = ; i < n; i++) {
int mi = a[i][];
for (int j = ; j < m; j++) mi = min(mi, a[i][j]);
r[i] = mi;
ans += r[i];
for (int j = ; j < m; j++) a[i][j] -= mi;
}
} void col_man() {
for (int j = ; j < m; j++) {
int mi = a[][j];
for (int i = ; i < n; i++) mi = min(mi, a[i][j]);
c[j] = mi;
ans += c[j];
for (int i = ; i < n; i++) a[i][j] -= mi;
}
} int main() {
cin >> n >> m;
for (int i = ; i < n; i++)
for (int j = ; j < m; j++)
scanf("%d", &a[i][j]);
if (n > m) {
col_man();
row_man();
} else {
row_man();
col_man();
}
if (!check()) puts("-1");
else {
cout << ans << endl;
for (int i = ; i < n; i++)
for (int j = ; j < r[i]; j++)
printf("row %d\n", i + );
for (int i = ; i < m; i++)
for (int j = ; j < c[i]; j++)
printf("col %d\n", i + );
}
}