Karen and Game is the third article about Karen. In a row. Pretty much, someone in CodeForces has created 7 different problems with her, thus the Japanese lady is getting famous.
This time the problem looks like this:
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!
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).
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.
And the solution has costed me really a lot of time. Mainly because I was naive and I was thinking that concatenating a string and printing it as an answer is faster than simply iterating through a list with the answers. Theoretically, we print the string once and with the list it could be up to 500 times. Still, 500 times printing from a list was faster (see the commented lines in the solution).
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using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; class Startup { public static int intResult = 0; //public static string strResult = ""; public static List<string> liResult = new List<string>(); static void Main() { int[] intInput = ReadLineAndParseToArray(); int x = intInput[0]; int y = intInput[1]; int[][] arrMatrix = new int[x][]; for (int i = 0; i < x; i++) arrMatrix[i] = ReadLineAndParseToArray(); if (y>x) { CheckRows(x, y, arrMatrix); CheckColumns(x, y, arrMatrix); } else { CheckColumns(x, y, arrMatrix); CheckRows(x, y, arrMatrix); } if (SumOfNestedArray(arrMatrix)!=0) { Console.WriteLine(-1); } else { Console.WriteLine(intResult); for (int i = 0; i < liResult.Count; i++) { Console.WriteLine(liResult[i]); } //Console.WriteLine(strResult); } } public static void CheckRows(int x, int y, int[][] arrMatrix) { for (int i = 0; i < x; i++) { int intMin = GetMinValueOfArray(arrMatrix[i]); if (intMin != 0) arrMatrix[i] = IncrementDecrementArray(arrMatrix[i], intMin * -1); for (int p = 0; p < intMin; p++) liResult.Add (String.Format("row {0}\n", i + 1)); //strResult += String.Format("row {0}\n", i + 1); intResult += intMin; } } public static void CheckColumns(int x,int y, int[][] arrMatrix) { for (int col = 0; col < y; col++) { int intMin = GetMinValueOfArrayHorizontal(arrMatrix, col); if (intMin != 0) { for (int row = 0; row < arrMatrix.Count(); row++) { arrMatrix[row][col]-=intMin; } for (int p = 0; p < intMin; p++) { liResult.Add(String.Format("col {0}\n", col + 1)); //strResult += String.Format("col {0}\n", col + 1); } intResult += intMin; } } } public static int SumOfNestedArray(int[][] arrMatrix) { int intResult = 0; for (int i = 0; i < arrMatrix.Count(); i++) { intResult += arrMatrix[i].Sum(); } return intResult; } public static int GetMinValueOfArrayHorizontal (int[][] intArray, int intPosition) { int intMinValue = int.MaxValue; for (int i = 0; i < intArray.Count(); i++) { intMinValue = Math.Min(intMinValue, intArray[i][intPosition]); } return intMinValue; } public static int GetMinValueOfArray(int[] intArray) { int intMinValue = int.MaxValue; for (int i = 0; i < intArray.Count(); i++) { intMinValue = Math.Min(intMinValue, intArray[i]); } return intMinValue; } public static int[] IncrementDecrementArray(int[] myArray, int intIncrementDecrement) { for (int i = 0; i < myArray.Count(); i++) { myArray[i] += intIncrementDecrement; } return myArray; } public static List<int> IncrementDecrementList(List<int> myList, int intValueToDecrement) { List<int> myListResult = new List<int>(myList.Count); for (int i = 0; i < myList.Count; i++) { myListResult.Add(myList[i] + intValueToDecrement); } return myListResult; } public static List<List<T>> TransposeMyList<T>(List<List<T>> lists) { var longest = lists.Any() ? lists.Max(l => l.Count) : 0; List<List<T>> outer = new List<List<T>>(longest); for (int i = 0; i < longest; i++) outer.Add(new List<T>(lists.Count)); for (int j = 0; j < lists.Count; j++) for (int i = 0; i < longest; i++) outer[i].Add(lists[j].Count > i ? lists[j][i] : default(T)); return outer; } public static int[,] TransposeMaxtrix(int[,] matrix) { int w = matrix.GetLength(0); int h = matrix.GetLength(1); int[,] result = new int[h, w]; for (int i = 0; i < w; i++) { for (int j = 0; j < h; j++) { result[j, i] = matrix[i, j]; } } return result; } public static int[] ReadLineAndParseToArray() { return Console.ReadLine().Split().Select(int.Parse).ToArray(); } public static List<int> ReadLineAndParseToList() { return Console.ReadLine().Split().Select(int.Parse).ToList(); } } |
What is the solution about? Pretty much, it is not that difficult – we check what do we have more – columns or rows. If we have more columns, it’s obvious that we would like to switch rows first. And vice versa. Then we check the minimal value in the line. If it is not zero, we decrement the values of the line with it. At the end we check the sum of the whole matrix. If it is zero, we print the values in the liResult. And that’s it. We have a matrix, which makes it easy to iterate columns and rows. Initially I was trying with list of lists, but the iteration process was a bit overcomplicated.
Yup, that’s all! Cheers! 🙂