Codeforces Round #419 (Div. 2) C. Karen and Game-优快云博客

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 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

output

4
row 1
row 1
col 4
row 3

input

output

-1

input

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:

$\text{[math]}$

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.

因为是输出任意一组，而且只能+1，所以倒着每行每列考虑，看n和m谁小，若n小先把行处理完在处理列，反之同理。

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#define inf 99999999

using namespace std;
typedef long long ll;
const int maxn = 100 + 10;
int a[maxn][maxn];
struct node {
	string s;
	int x;
	node(string ss, int xx): s(ss), x(xx) {}
};
vector<node> v;
int n, m;
int main()
{
	while (cin >> n >> m) {
		v.clear();
		bool flag = false;
		for (int i = 1 ; i <= n ; i ++) {
			for (int j = 1 ; j <= m ; j ++) {
				cin >> a[i][j];
				if (a[i][j] != 0)flag = true;
			}
		}
		if (flag == false) {cout << "0" << endl; continue;}
		for (int i = 1 ; i <= n ; i++) {
			a[i][0] = inf;
			for (int j = 1 ; j <= m ; j ++) {
				a[i][0] = min(a[i][0], a[i][j]);
			}
		}
		for (int j = 1 ; j <= m ; j ++) {
			a[0][j] = inf;
			for (int i = 1 ; i <= n ; i ++) {
				a[0][j] = min(a[0][j], a[i][j]);
			}
		}
		if (n <= m) {
			for (int i = 1 ; i <= n ; i++) {
				if (a[i][0] == 0)continue;
				for (int k = a[i][0] ; k >= 1 ; k --) {
					v.push_back(node("row", i));
				}
				for (int j = 1 ; j <= m ; j ++) {
					a[0][j] = min(a[0][j], a[i][j] - a[i][0]);
				}
				for (int j = 1 ; j <= m ; j ++) {
					a[i][j] -= a[i][0];
				}
			}
			for (int j = 1 ; j <= m ; j++) {
				if (!a[0][j])continue;
				for (int k = 1 ; k <= a[0][j] ; k++) {
					v.push_back(node("col", j));
				}
				for (int i = 1 ; i <= n ; i ++) {
					a[i][j] -= a[0][j];
				}
			}
		}
		else {
			for (int j = 1 ; j <= m ; j++) {
				if (!a[0][j])continue;
				for (int k = 1 ; k <= a[0][j] ; k++) {
					v.push_back(node("col", j));
					//cout << "col" << " " << j << endl;
				}
				for(int i = 1 ; i <= n ; i ++){
					a[i][0] = min(a[i][0],a[i][j]-a[0][j]);
				}
				for (int i = 1 ; i <= n ; i ++) {
					a[i][j] -= a[0][j];
				}
				
			}
			for (int i = 1 ; i <= n ; i++) {
				if (a[i][0] == 0)continue;
				for (int k = a[i][0] ; k >= 1 ; k --) {
					v.push_back(node("row", i));
				}
				for (int j = 1 ; j <= m ; j ++) {
					a[i][j] -= a[i][0];
				}
			}
		}
		bool ok = false;
		for (int i = 1 ; i <= n ; i ++) {
			for (int j = 1 ; j <= m ; j ++) {
				if (a[i][j] != 0)ok = true;
			}
		}
		/*for(int i = 1 ; i <= n ; i ++){
			for(int j = 1 ; j <= m ; j ++){
				cout << a[i][j] << " ";
			}
			cout << endl;
		}*/
		if (ok) {cout << "-1" << endl; continue;}
		if (v.size() == 0 && flag == true) {cout << "-1" << endl; continue;}
		cout << v.size() << endl;
		for (int i = v.size() - 1 ; i >= 0 ; i --) {
			cout << v[i].s << " " << v[i].x << endl;
		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/HsiaoYeekwan/p/7043908.html