M文件

function [intx,intf] = ZeroOneprog(c,A,b,x0) %目标函数系数向量，c %不等式约束矩阵，A %不等式约束右端向量，b %初始整数可行解，x0

%目标函数取最小值时的自变量值，intx %目标函数的最小值，intf

sz = size(A); if sz(2) < 3

[intx,intf] = Allprog(c,A,b); %穷举法 else

[intx,intf] = Implicitprog(c,A,b,x0); %隐枚举法 end

function [intx,intf] = Allprog(c,A,b); sz_A = size(A); rw = sz_A(1); col = sz_A(2);

minf = inf;

for i=0:(2^(col)-1) %枚举空间

x1 = myDec2Bin(i,col); %十进制转化为二进制 if A*x1 >= b %是否满足约束条件 f_tmp = c*x1; if f_tmp < minf minf = f_tmp; intx = x1; intf = minf; else

continue; end else

continue; end end

function [intx,intf] = Implicitprog(c,A,b,x0)%隐枚举法 sz_A = size(A); rw = sz_A(1); col = sz_A(2);

minf = c*x0; A = [A;-c];

b = [b;-minf]; %增加了一个限制分量 for i=0:(2^(col)-1)

x1 = myDec2Bin(i,col); if A*x1 >= b

f_tmp = c*x1; if f_tmp < minf minf = f_tmp;

b(rw+1,1) = -minf; %隐枚举法与穷举法的区别在于此句 intx = x1; intf = minf; else

continue; end else

continue; end end

function y = myDec2Bin(x,n) %十进制转化为二进制 str = dec2bin(x,n); for j=1:n

y(j) = str2num(str(j)); end

y = transpose(y);

求解实例

求解下面0-1规划

?2x1?3x2?5x3?4x4?7x5?8?minf?x??x1?2x2?3x3?x4?x5,s.t.?x1?2x2?4x3?2x4?2x5?5

?x,x,x,x,x?0或1?12345在MATLAB命令框在输入下列命令：

>> c=[1 2 3 1 1];

>> A=[2 3 5 4 7;1 1 4 2 2]; >> b=[8;5];

>> x0=[1;1;1;1;1]’;

>> [intx,intf]=ZetoOneprog(c,A,b,x0)

所得结果如下：

intx =

1 0 0 1 1

intf =

3