Question

首先，感谢所有读者。

以下问题。

    keys = pygame.key.get_pressed()

    if pl_rect.colliderect(obs1_rect_left):
        col_left = True
    else:
        col_left = False

    if pl_rect.colliderect(obs1_rect_right):
        col_right = True
    else:
        col_right = False

    if pl_rect.colliderect(obs1_rect_down):
        col_down = True
    else:
        col_down = False
    if pl_rect.colliderect(obs1_rect_up):
        col_up = True
    else:
        col_up = False




    if keys[pygame.K_w] and not(col_up):
        pl.y -= pl.vel

    if keys[pygame.K_s] and not(col_down):
        pl.y += pl.vel

    if keys[pygame.K_d] and not(col_right):
        pl.x += pl.vel

    if keys[pygame.K_a] and not(col_left):
        pl.x -= pl.vel

在这种情况下，我想通过选择某些低点和列来重新构建矩阵。

就像...

int MaxANum = 5;
int MaxBNum = 3;
int MaxENum = 8;

int ANum = 3;
int BNum = 2;
int ENum = 5;

range TTRange = 1..(MaxANum+MaxBNum);
range TRange = 1..(ANum+BNum);
range ARange = 1..ANum;
range BRange = 1..BNum;

range TERange = 1..MaxENum;
range ERange = 1..ENum;

M[TTRange][TERange] = [[**0, 1, 1, 0, 1**, 1, 0, 1] -> 1st ANum
                    [**1, 1, 0, 0, 1**, 0, 0, 1] -> 2nd ANum
                    [**0, 0, 1, 0, 1**, 0, 1, 0] -> 3rd ANum
                    [0, 1, 0, 0, 1, 1, 1, 0] -> 4th ANum
                    [1, 1, 0, 1, 0, 1, 0, 1] -> 5th ANum
                    [**1, 0, 0, 1, 1**, 0, 0, 0] -> 1st BNum
                    [**0, 0, 1, 0, 0, 0**, 1, 1] -> 2nd BNum
                    [1, 1, 0, 0, 0, 0, 0, 1]]; -> 3rd BNum

我是CPLEX和计算机编码的初学者。请告诉我如何获取此代码。非常感谢您的阅读。

Answer 1

我不太确定您的输入与所需输出的关系如何，但是此代码段可能会有所帮助：

range TTRange = 1..8;
range TRange = 1..5;
range TERange = 1..8;
range ERange = 1..5;

int M[TTRange][TERange] = [[0, 1, 1, 0, 1, 1, 0, 1],
                           [1, 1, 0, 0, 1, 0, 0, 1],
                           [0, 0, 1, 0, 1, 0, 1, 0],
                           [0, 1, 0, 0, 1, 1, 1, 0],
                           [1, 1, 0, 1, 0, 1, 0, 1],
                           [1, 0, 0, 1, 1, 0, 0, 0],
                           [0, 0, 1, 0, 0, 0, 1, 1],
                           [1, 1, 0, 0, 0, 0, 0, 1]];

int M1[t in TRange][e in ERange] = M[t][e];

它将生成此子矩阵：

 M1 = [[0 1 1 0 1]
       [1 1 0 0 1]
       [0 0 1 0 1]
       [0 1 0 0 1]
       [1 1 0 1 0]]

“技巧”是在M1的索引集中进行选择。

如何对矩阵进行变形和变形？

1 个答案: