[刷题]Codeforces 815A - Karen and Game

千年两小时只写出2题，这次终于把C题写出来了。。。

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.

Key

有一个有n*m个单元的、每个单元填有数字的网格，每次操作可将某一整行、一整列的单元数值-1，问最少进行多少次操作可以将网格所有单元减到0。

模拟。对每一行/列模拟即可。对于3*5的网格，如果要把每一个单元的数值-1，按行减，要3次，按列减，要5次。所以模拟的时候肯定是优先按行数/列数小的那一个开始减，每次减到该行/列出现0则无法再减，跳转下一行/列。模拟到最后遍历一遍查看有没有个别单元无法减到0，若有则输出-1。

Code

#include<iostream>
#include<cstring>
using namespace std;
int n, m;
int arr[123][123];
int minr[123], minc[123];
int main()
{
    ios::sync_with_stdio(false);
    memset(minr, 1, sizeof(minr));
    memset(minc, 1, sizeof(minc));
    cin >> n >> m;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            int &now = arr[i][j];
            cin >> now;
            if (minr[i] > now) minr[i] = now;
            if (minc[j] > now) minc[j] = now;
        }
    }
    int res = 0;
    if (n <= m) {
        int rminr = 9999999;
        for (int i = 0; i < n; ++i) {
            res += minr[i];
            if (rminr > minr[i]) rminr = minr[i];
        }
        for (int i = 0; i < m; ++i) {
            res += (minc[i] -= rminr);
        }
    }
    else {
        int cminc = 9999999;
        for (int i = 0; i < m; ++i) {
            res += minc[i];
            if (cminc > minc[i]) cminc = minc[i];
        }
        for (int i = 0; i < n; ++i) {
            res += (minr[i] -= cminc);
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (arr[i][j] - minr[i] - minc[j]) {
                cout << "-1\n";
                return 0;
            }
        }
    }
    cout << res << '\n';
    for (int i = 0; i < n; ++i) {
        while (minr[i]--) cout << "row " << (i + 1) << '\n';
    }
    for (int i = 0; i < m; ++i) {
        while (minc[i]--) cout << "col " << (i + 1) << '\n';
    }
    return 0;
}