在我的 Minizinc 项目中,我试图生成一个包含 n 组的数组。
给定一个由 t 个不同数字组成的数组,生成 n 个不同的集合,其
基数在数组 m 中给出。
例如:
t=10; n = 4; m = [3, 2, 2, 3];
我想生成一个集合数组 x = [1..3, 4..5, 6..7, 8..10];
但是我从下面的代码中得到的是
x = [1..3, 4..5, {6,10}, 7..9];
(我不想使用求解最小化或其他各种求解作为我的
目的只是生成中间的集合数组。)
int: n = 4; % number of groups
array[1..n] of int: m = [3, 2, 2, 3]; % size of each group
int: t = sum(i in 1..n)(m[i]); % total members
array[1..n] of var set of 1..t: x; % the array of sets
constraint forall(i in 1..n-1)(x[i] > x[i+1]); % SORT .
constraint forall(i in 1..n)(card(x[i] ) = m[i]); % Size of each set
constraint forall(i in 1..n-1)( x[i] intersect x[i+1] = {}); %
% I can't see a way to keep the digits in order
%constraint array_intersect(x) = {}; % this didn't help
solve satisfy;
output [show(x)];
最佳答案
您可以毫无限制地执行此操作。这是一种方法,虽然有点难看:
int: n = 4; % number of sets
array[1..n] of int: s = [3,2,2,3]; % cardinality of the sets
array[1..n] of set of int: x = [ {k | k in sum([s[j] | j in 1..i-1])+1..sum([s[j] | j in 1..i]) } | i in 1..n];
solve satisfy;
constraint true ; % just used to run the model
output [ "x: \(x)\n"];
关于arrays - Minizinc 阵列套装,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44644128/