Question

我正试图创造一种“你沉没我的战舰”游戏。到目前为止，我有两个数组： 1.板（表） - 要播放的表面 2.船（Avion） - 在这种情况下是飞机

我想要实现的是创建一个包含整个表面并将平面放在其中的第三个数组。

所以我有飞机：

和表面：

  1 2 3 4 5 6 7 8 8 10
A 0 0 0 0 0 0 0 0 0 0
B 0 0 0 0 0 0 0 0 0 0
C 0 0 0 0 0 0 0 0 0 0
D 0 0 0 0 0 0 0 0 0 0
E 0 0 0 0 0 0 0 0 0 0
F 0 0 0 0 0 0 0 0 0 0
G 0 0 0 0 0 0 0 0 0 0
H 0 0 0 0 0 0 0 0 0 0
I 0 0 0 0 0 0 0 0 0 0
J 0 0 0 0 0 0 0 0 0 0

我想要达到的目标是：

  1 2 3 4 5 6 7 8 8 10
A 0 0 1 0 0 0 0 0 0 0
B 1 1 1 1 1 0 0 0 0 0
C 0 0 1 0 0 0 0 0 0 0
D 0 0 1 0 0 0 0 0 0 0
E 0 1 1 1 0 0 0 0 0 0
F 0 0 0 0 0 0 0 0 0 0
G 0 0 0 0 0 0 0 0 0 0
H 0 0 0 0 0 0 0 0 0 0
I 0 0 0 0 0 0 0 0 0 0
J 0 0 0 0 0 0 0 0 0 0

int counterAvion = 0;
        int counterTable = 0;

        int[,] Table = new int[10, 10];
        int[,] Avion = new int[5, 5];
        int[,] Combined = new int[10, 10];


        // Table[][]
        //rowA
        Table[0, 0] = 0; Table[0, 1] = 0; Table[0, 2] = 0; Table[0, 3] = 0; Table[0, 4] = 0;
        Table[0, 5] = 0; Table[0, 6] = 0; Table[0, 7] = 0; Table[0, 8] = 0; Table[0, 9] = 0;
        //rowB
        Table[1, 0] = 0; Table[1, 1] = 0; Table[1, 2] = 0; Table[1, 3] = 0; Table[1, 4] = 0;
        Table[1, 5] = 0; Table[1, 6] = 0; Table[1, 7] = 0; Table[1, 8] = 0; Table[1, 9] = 0;
        //rowC
        Table[2, 0] = 0; Table[2, 1] = 0; Table[2, 2] = 0; Table[2, 3] = 0; Table[2, 4] = 0;
        Table[2, 5] = 0; Table[2, 6] = 0; Table[2, 7] = 0; Table[2, 8] = 0; Table[2, 9] = 0;
        //rowD
        Table[3, 0] = 0; Table[3, 1] = 0; Table[3, 2] = 0; Table[3, 3] = 0; Table[3, 4] = 0;
        Table[3, 5] = 0; Table[3, 6] = 0; Table[3, 7] = 0; Table[3, 8] = 0; Table[3, 9] = 0;
        //rowE
        Table[4, 0] = 0; Table[4, 1] = 0; Table[4, 2] = 0; Table[4, 3] = 0; Table[4, 4] = 0;
        Table[4, 5] = 0; Table[4, 6] = 0; Table[4, 7] = 0; Table[4, 8] = 0; Table[4, 9] = 0;
        //rowF
        Table[5, 0] = 0; Table[5, 1] = 0; Table[5, 2] = 0; Table[5, 3] = 0; Table[5, 4] = 0;
        Table[5, 5] = 0; Table[5, 6] = 0; Table[5, 7] = 0; Table[5, 8] = 0; Table[5, 9] = 0;
        //rowG
        Table[6, 0] = 0; Table[6, 1] = 0; Table[6, 2] = 0; Table[6, 3] = 0; Table[6, 4] = 0;
        Table[6, 5] = 0; Table[6, 6] = 0; Table[6, 7] = 0; Table[6, 8] = 0; Table[6, 9] = 0;
        //rowH
        Table[7, 0] = 0; Table[7, 1] = 0; Table[7, 2] = 0; Table[7, 3] = 0; Table[7, 4] = 0;
        Table[7, 5] = 0; Table[7, 6] = 0; Table[7, 7] = 0; Table[7, 8] = 0; Table[7, 9] = 0;
        //rowI
        Table[8, 0] = 0; Table[8, 1] = 0; Table[8, 2] = 0; Table[8, 3] = 0; Table[8, 4] = 0;
        Table[8, 5] = 0; Table[8, 6] = 0; Table[8, 7] = 0; Table[8, 8] = 0; Table[8, 9] = 0;
        //rowJ
        Table[9, 0] = 0; Table[9, 1] = 0; Table[9, 2] = 0; Table[9, 3] = 0; Table[9, 4] = 0;
        Table[9, 5] = 0; Table[9, 6] = 0; Table[9, 7] = 0; Table[9, 8] = 0; Table[9, 9] = 0;

        // Avion[][]
        //head
        Avion[0, 0] = 0; Avion[0, 1] = 0; Avion[0, 2] = 1; Avion[0, 3] = 0; Avion[0, 4] = 0;
        //wings
        Avion[1, 0] = 1; Avion[1, 1] = 1; Avion[1, 2] = 1; Avion[1, 3] = 1; Avion[1, 4] = 1;
        //mid_1
        Avion[2, 0] = 0; Avion[2, 1] = 0; Avion[2, 2] = 1; Avion[2, 3] = 0; Avion[2, 4] = 0;
        //mid_2
        Avion[3, 0] = 0; Avion[3, 1] = 0; Avion[3, 2] = 1; Avion[3, 3] = 0; Avion[3, 4] = 0;
        //tail
        Avion[4, 0] = 0; Avion[4, 1] = 1; Avion[4, 2] = 1; Avion[4, 3] = 1; Avion[4, 4] = 0;



        // Output Avion[][]
        Console.WriteLine("\nAirplane:\n");
        foreach (var item in Avion)
        {
            counterAvion++;
            Console.Write(item.ToString() + "  ");
            if (counterAvion == 5)
            {
                Console.WriteLine("\n");
                counterAvion = 0;
            }
        }

        // Output Table[][]
        Console.WriteLine("\nTable:\n");
        Console.WriteLine("  1  2  3  4  5  6  7  8  9  10\n");
        int counterColumn = 1;
        foreach (var item in Table)
        {
            counterTable++;
            //Console.WriteLine(counterColumn);
            switch (counterColumn)
            {
                case 1:
                    Console.Write("A ");
                    break;
                case 12:
                    Console.Write("B ");
                    break;
                case 23:
                    Console.Write("C ");
                    break;
                case 34:
                    Console.Write("D ");
                    break;
                case 45:
                    Console.Write("E ");
                    break;
                case 56:
                    Console.Write("F ");
                    break;
                case 67:
                    Console.Write("G ");
                    break;
                case 78:
                    Console.Write("H ");
                    break;
                case 89:
                    Console.Write("I ");
                    break;
                case 100:
                    Console.Write("J ");
                    break;
                default:
                    break;
            }

            Console.Write(item.ToString() + "  ");
            if (counterTable == 10)
            {
                Console.WriteLine("\n");
                counterTable = 0;
                counterColumn++;
            }
            counterColumn++;
        }
Console.ReadLine();

我是C＃的新手，到目前为止，我喜欢它，但它是......嗯...不同于我很久以前玩过的BorlandC。

那么，关于如何实现这一点的任何想法？

我也知道到目前为止我的代码有点混乱，所以对此的任何意见都表示赞赏。

Answer 1

复制值并不是特别困难。如果你想创建第三个数组，你可以在开始时创建第一个数组的直接副本，然后修改它。

public void Combine(int[,] first, int[,] second, int rowStart, int colStart)
{
    //TODO validate parameters, ensure arrays are all large enough, etc.

    for (int i = 0; i < second.GetLength(0); i++)
    {
        for (int j = 0; j < second.GetLength(1); j++)
        {
            first[colStart + i, rowStart + j] |= second[i, j];
        }
    }
}

Answer 2

假设飞机的尺寸永远不会超过表面，您可以直接穿过飞机，对于每个索引，您将表面设置为相同的值。

for(int i = 0; i < Airplane.length(0); i++){
    for(int j = 0; j < Airplane.length(1); j++){         
        surface[i,j] = Airplane[i,j]

如果不是，您可以轻松计算偏移量。

c＃分层两个数组

2 个答案: