Problem:
Score : 400 points

Problem Statement

There is a grid of square cells with H horizontal rows and W vertical columns. The cell at the i-th row and the j-th column will be denoted as Cell (i, j).

In Cell (i, j), a_{ij} coins are placed.

You can perform the following operation any number of times:

Operation: Choose a cell that was not chosen before and contains one or more coins, then move one of those coins to a vertically or horizontally adjacent cell.

Maximize the number of cells containing an even number of coins.

Constraints

All values in input are integers.

1 \leq H, W \leq 500

0 \leq a_{ij} \leq 9

Input

Input is given from Standard Input in the following format:

H W
a_{11} a_{12} ... a_{1W}
a_{21} a_{22} ... a_{2W}
:
a_{H1} a_{H2} ... a_{HW}

Output

Print a sequence of operations that maximizes the number of cells containing an even number of coins, in the following format:

N
y_1 x_1 y_1' x_1'
y_2 x_2 y_2' x_2'
:
y_N x_N y_N' x_N'

That is, in the first line, print an integer N between 0 and H \times W (inclusive), representing the number of operations.

In the (i+1)-th line (1 \leq i \leq N), print four integers y_i, x_i, y_i' and x_i' (1 \leq y_i, y_i' \leq H and 1 \leq x_i, x_i' \leq W), representing the i-th operation. These four integers represents the operation of moving one of the coins placed in Cell (y_i, x_i) to a vertically or horizontally adjacent cell, (y_i', x_i').

Note that if the specified operation violates the specification in the problem statement or the output format is invalid, it will result in Wrong Answer.

Sample Input 1

2 3
1 2 3
0 1 1

Sample Output 1

3
2 2 2 3
1 1 1 2
1 3 1 2

Every cell contains an even number of coins after the following sequence of operations:

Move the coin in Cell (2, 2) to Cell (2, 3).

Move the coin in Cell (1, 1) to Cell (1, 2).

Move one of the coins in Cell (1, 3) to Cell (1, 2).

Sample Input 2

3 2
1 0
2 1
1 0

Sample Output 2

3
1 1 1 2
1 2 2 2
3 1 3 2

Sample Input 3

1 5
9 9 9 9 9

Sample Output 3

2
1 1 1 2
1 3 1 4

Solution in Scala (2.11.7) (judge verdict: Accepted)

import scala.collection.immutable.Queue

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val H, W = sc.nextInt()
    val A = Array.fill(H,W)(sc.nextInt())

    // １マスにつき一回、１枚だけコインを上下左右のどこかのマスに移動させれる
    // 偶数枚のコインが置かれたマスの数を最大化
    // 既に偶数枚置かれているマスは, 動かしても意味がない
    // 奇数枚置かれているマスから奇数枚置かれているマスに移動させるのがベスト
    // ただし, 奇数枚置かれているマスから偶数枚動かすパターンでもあとあと全部偶数にできるケースがある
    // 奇数枚のマスと奇数枚のマスが離れていて, その間の偶数枚のマスの上をコインをスライドさせれば
    // その経路は使用不可能になるが全て偶数枚のマスにすることができる
    // 奇数枚のマスから奇数枚のマスへのBFSを行えば良さそう
    // すると, 奇数枚のマスを発見した場合, そこから最短で見つかる奇数枚のマスを見つける
    // 経路を覚えておき, 使った経路のみを次回の探索に使えないようにし, かつ操作として加える,
    type Path = (Int, Int, Int, Int)
//    def solve(H: Int, W: Int, A: Array[Array[Int]]): List[Path] = {
//      val dx = Array(-1, 0, 1, 0)
//      val dy = Array(0, -1, 0, 1)
//      val used = Array.fill(H, W)(false)
//      val visited = Array.fill(H, W)(false)
//      def clear(): Unit =
//        for {
//          i <- 0 until H
//          j <- 0 until W
//        } visited(i)(j) = false
//
//      def find(sx: Int, sy: Int): Option[List[Path]] = {
//        def bfs(queue: Queue[((Int, Int), List[Path])]): Option[List[Path]] =
//          if (queue.nonEmpty) {
//            val (((x, y), _path), next) = queue.dequeue
//            if (!(x == sx && y == sy) && A(y)(x) % 2 == 1) {
//              _path foreach { case (x, y, gx, gy) =>
//                used(y)(x)   = true
//                used(gy)(gx) = true
//              }
//              Some(_path)
//            } else {
//              bfs((0 until 3).foldLeft(next)((queue, i) => {
//                val nx = x + dx(i)
//                val ny = y + dy(i)
//                if (nx >= 0 && nx < W && ny >= 0 && ny < H && !used(ny)(nx) && !visited(ny)(nx)) {
//                  visited(ny)(nx) = true
//                  queue.enqueue(((nx, ny), (x, y, nx, ny) :: _path))
//                } else queue
//              }))
//            }
//          } else None
//        bfs(Queue(((sx, sy), List[Path]())))
//      }
//
//      var paths = List[Path]()
//      for {
//        i <- 0 until H
//        j <- 0 until W
//      } yield {
//        if (!used(i)(j) && A(i)(j) % 2 == 1) {
//          find(j, i) foreach { p =>
//            paths = p ++ paths
//          }
//          clear()
//        }
//      }
//      paths.reverse
//    }

    // 実際には、一筆書きをすればよかった
    // 奇数から奇数にスライドすれば偶数になるので,
    // 奇数から始まり奇数に終わる全てのマスについて, 一筆書きを行うと
    // 奇数マスが偶数個あれば, 全てのマスについて偶数にできるが
    // 奇数個の場合には一つだけ残る
    // また, 無駄な操作があっても良い
    def solve(H: Int, W: Int, A: Array[Array[Int]]): List[Path] = {
      def moveRight(x: Int, y: Int, path: List[Path]): List[Path] = {
        if (x < W - 1) {
          if (A(y)(x) % 2 == 1) {
            A(y)(x) -= 1
            A(y)(x+1) += 1
            moveRight(x + 1, y, (x, y, x+1, y) :: path)
          } else moveRight(x + 1, y, path)
        } else {
          if (A(y)(x) % 2 == 1) {
            if (y < H - 1) {
              A(y)(x) -= 1
              A(y+1)(x) += 1
              moveLeft(x, y + 1, (x, y, x, y + 1) :: path)
            } else path
          } else {
            if (y < H - 1) {
              moveLeft(x, y + 1, path)
            } else path
          }
        }
      }
      def moveLeft(x: Int, y: Int, path: List[Path]): List[Path] = {
        if (x > 0) {
          if (A(y)(x) % 2 == 1) {
            A(y)(x) -= 1
            A(y)(x-1) += 1
            moveLeft(x - 1, y, (x, y, x-1, y) :: path)
          } else moveLeft(x - 1, y, path)
        } else {
          if (A(y)(x) % 2 == 1) {
            if (y < H - 1) {
              A(y)(x) -= 1
              A(y + 1)(x) += 1
              moveRight(x, y + 1, (x, y, x, y + 1) :: path)
            } else path
          } else {
            if (y < H - 1) {
              moveRight(x, y + 1, path)
            } else path
          }
        }
      }
      moveRight(0, 0, Nil).reverse
    }

    val paths = solve(H, W, A)
    println(paths.size)
    println(paths.map { case (x, y, gx, gy) =>
      s"${y+1} ${x+1} ${gy+1} ${gx+1}"
    }. mkString("\n"))
  }

}