【#文档大全网# 导语】以下是®文档大全网的小编为您整理的《匈牙利算法MATLAB程序》，欢迎阅读！
%%用匈牙利法解指派问题
%%C是效益矩阵
function [y,fval,flag]=Hungary(C)
[m,n]=size(C);
tempC=C;
%%%以下两个循环使每行每列都出现零元素
for i=1:m
tempC(i,:)=tempC(i,:)-min(tempC(i,:));
end
for i=1:n
tempC(:,i)=tempC(:,i)-min(tempC(:,i));
end
AssignMatrix=zeros(m,n);
tempC=TryAssign(tempC);
OneNumber=0;
for i=1:m
for j=1:n
if tempC(i,j)==inf
OneNumber=OneNumber+1;
break;
end
end
end
while OneNumber<m
Row=zeros(m,1);
Col=ones(1,n);
Line=[];
for i=1:m
if IsInMatrix(inf,tempC(i,:))==0
Line=[Line,i];
Row(i)=1;
end
end
for i=Line
Cur=i;
while Cur~=0
[Cur,Row,Col]=ZeroCover(tempC,Row,Col,Cur);
end
end
temp=inf;
for i=1:m
for j=1:n
if Row(i)==1&Col(j)==1&tempC(i,j)<temp
temp=tempC(i,j);
end
end
end
for i=1:m
for j=1:n
if tempC(i,j)==inf|tempC(i,j)==-inf
tempC(i,j)=0;
end
end
end
for i=1:m
if Row(i)==1
tempC(i,:)=tempC(i,:)-temp;
end
end
for j=1:n
if Col(j)==0
tempC(:,j)=tempC(:,j)+temp;
end
end
tempC=TryAssign(tempC);
OneNumber=0;
for i=1:m
for j=1:n
if tempC(i,j)==inf
OneNumber=OneNumber+1;
break;
end
end
end
end
for i=1:m
for j=1:n
if tempC(i,j)==inf
AssignMatrix(i,j)=1;
end
end
end
% y=AssignMatrix;
for i=1:m
for j=1:m
if AssignMatrix(i,j)==1
y(i,1)=i;
y(i,2)=j;
break;
end
end
end
col=zeros(1,m);
row=zeros(1,m);
for i=1:m
col=sum(AssignMatrix(i,:));
row=sum(AssignMatrix(:,i));
end
if col==ones(1,m)&row==ones(1,m)
flag=1;
else
flag=0;
end
temp=C.*AssignMatrix;
fval=sum(temp(:));
%% 对处理过的每行每列都有零元素的矩阵C进行试指派
function y=TryAssign(C)
[m,n]=size(C);
while IsInMatrix(0,C)==1
flag=0;
for i=1:m
if ZeroNumber(C(i,:))==1
temp=C(i,:);
ZeroRowPos=find(temp==0);
C(i,ZeroRowPos)=inf;
temp=C(:,ZeroRowPos);
ZeroColPos=find(temp==0);
for j=ZeroColPos
C(j,ZeroRowPos)=-inf;
end
flag=flag+1;
end
end
for i=1:n
if ZeroNumber(C(:,i))==1
temp=C(:,i);
ZeroColPos=find(temp==0);
C(ZeroColPos,i)=inf;
temp=C(ZeroColPos,:);
ZeroRowPos=find(temp==0);
for j=ZeroRowPos
C(ZeroColPos,j)=-inf;
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2022-10-09
