指派授课问题
现有A、B、C、D四门课程，需由甲、乙、丙、丁四人讲授，并且规定：
每人只讲且必须讲1门课；每门课必须且只需1人讲。
四人分别讲每门课的费用示于表中：
带包python代码：
from scipy.optimize import linear_sum_assignmentimport numpy as np#cost =np.array([[4,1,3],[2,0,5],[3,2,2]])cost =np.array([[2,10,9,7],[15,4,14,8],[13,14,16,11],[4,15,13,9]])row_ind,col_ind=linear_sum_assignment(cost)print(row_ind)#开销矩阵对应的行索引print(col_ind)#对应行索引的最优指派的列索引print(cost[row_ind,col_ind])#提取每个行索引的最优指派列索引所在的元素，形成数组print(cost[row_ind,col_ind].sum())#数组求和#输出指派矩阵p = np.zeros((4,4))p[row_ind,col_ind]=1 print(p)
暴力python代码：
# -*- coding: utf-8 -*-import numpy as npimport copyc=[2,10,9,7,15,4,14,8,13,14,16,11,4,15,13 ,9]c = np.array(c)c = c.reshape((4,4))all_p=[]class obj:def _init_(self):self.p=[]self.cost=0for i in range(4):for j in range(4):if j==i:continue for u in range(4):if u==i or u==j :continuefor v in range(4):if v==i or v==j or v==u:continuep = np.zeros((4,4))p[0,i]=p[1,j]=p[2,u]=p[3,v]=1 ans = obj()ans.p = copy.deepcopy(p)ans.cost = sum(sum(c*ans.p))all_p.append(ans)all_p.sort(key=lambda ans: ans.cost, reverse=False)print(all_p[0].p)print(all_p[0].cost)
我写的matlab：
clearC=[2 10 9 715 4 14 813 14 16 114 15 13 9];A = perms(1:4);%perm显示1，2，3，4四个数的全排列L = length(A)best=999best_mat=[]for i=1:La = zeros(4,4);b = A(i,:);%遍历全排列中的每一种 c = 1:4;a(sub2ind(size(a), b, c))=1;%a矩阵指定的位置赋值为1s = sum(sum(a.*C));%求出费用和if best>s %挑出最大的best_mat=a;best=s;endendbest_matbest
老师的matlab代码1：
clearn=4;A=perms(1:n);G=size(A); %24 4 size(A,1) 24 size(A,2) 24n0=G(1); %24B=[2,10,9,7;15,4,14,8;13,14,16,11;4,15,13,9];for n1=1:n0%C为第n1中排列情况下，费用的4个取值C(1)=B(1,A(n1,1));C(2)=B(2,A(n1,2));C(3)=B(3,A(n1,3));C(4)=B(4,A(n1,4));%D{n1}表示第n1种情况下的4个取值D{n1}=[C(1),C(2),C(3),C(4)];s(n1)=sum(D{n1});end%找到最小的，返回a为行左边，b为纵坐标，a=1,b=9[a,b]=find(s==min(s));K=A(b,:)
根据老师的代码改进我的代码：
clearC=[2109715414813141611415139];A = perms(1:4);%perm显示1，2，3，4四个数的全排列L = length(A)for i=1:La = zeros(4,4);b = A(i,:);%遍历全排列中的每一种 c = 1:4;a(sub2ind(size(a), b, c))=1;%a矩阵指定的位置赋值为1D{i}=a;S(i)=sum(sum(a.*C));%求出费用和end[a,b]=find(S==min(S))D{b}S(b)
老师的matlab代码2：随机生成，不是很好，看运气
clearA=[2 15 13 4];B=[10 4 14 15];C=[9 14 16 13];D=[7 8 11 9];Y=zeros(1,1000);s=64;x=zeros(1,4);for i= 1:1000X=randperm(4);Y(i)=A(X(1))+B(X(2))+C(X(3))+D(X(4));if Y(i)<ss=Y(i);x=X;endend s,x
网上常见的matlab代码：
%适用于任意n阶系数矩阵clear all；C=[21097,154148,13141611,415139,];%效率矩阵Cn=size(C,1);%计算C的行列数nC=C(:);%计算目标函数系数，将矩阵C按列排成一个列向量即可。A=[];B=[];%没有不等式约束Ae=zeros(2*n,n^2);%计算等约束的系数矩阵afor i=1:nfor j=(i-1)*n+1:n*iAe(i,j)=1;endfor k=i:n:n^2Ae(n+i,k)=1;endendBe=ones(2*n,1);%等式约束右端项bXm=zeros(n^2,1);%决策变量下界XmXM=ones(n^2,1);%决策变量上界XM[x,z]=linprog(C,A,B,Ae,Be,Xm,XM);%使用linprog求解x=reshape(x,n,n);%将列向量x按列排成一个n阶方阵disp('最优解矩阵为:');%输出指派方案和最优值Assignment=round(x)%使用round进行四舍五入取整disp('最优解为:');z
线性规划matlab代码：
%线性规划c=[2,10,9,7,15,4,14,8,13,14,16,11,4,15,13,9];Aeq=[1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0;0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0;0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0;0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1;1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0;0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0;0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0;0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1];beq=[1,1,1,1,1,1,1,1];lb=zeros(16,1);ub=ones(16,1);[x,fval] = linprog(c,[],[],Aeq,beq,lb,ub)
