MATLAB实例:Munkres指派算法
作者:凯鲁嘎吉 - 博客园 http://www.cnblogs.com/kailugaji/
1. 指派问题陈述
指派问题涉及将机器分配给任务,将工人分配给工作,将足球运动员分配给职位等。目标是确定最佳分配,例如,使总成本最小化或使团队效率最大化。指派问题是组合优化领域中的一个基本问题。
例如,假设我们有四个工作需要由四个工作人员执行。因为每个工人都有不同的技能,所以执行工作所需的时间取决于分配给该工人的工人。
下面的矩阵显示了工人和工作的每种组合所需的时间(以分钟为单位)。作业用J1,J2,J3和J4表示,工人用W1,W2,W3和W4表示。
J1 | J2 | J3 | J4 | |
W1 | 82 | 83 | 69 | 92 |
W2 | 77 | 37 | 49 | 92 |
W3 | 11 | 69 | 5 | 86 |
W4 | 8 | 9 | 98 | 23 |
每个工人应仅执行一项工作,目标是最大程度地减少执行所有工作所需的总时间。
事实证明,将工人1分配给作业3,将工人2分配给作业2,将工人3分配给作业1,将工人4分配给作业4是最佳选择。那么,总时间为69 + 37 + 11 + 23 = 140分钟。所有其他任务导致需要更多时间。
2. Munkres指派算法MATLAB程序
munkres.m
function [assignment,cost] = munkres(costMat) % MUNKRES Munkres Assign Algorithm % % [ASSIGN,COST] = munkres(COSTMAT) returns the optimal assignment in ASSIGN % with the minimum COST based on the assignment problem represented by the % COSTMAT, where the (i,j)th element represents the cost to assign the jth % job to the ith worker. % % This is vectorized implementation of the algorithm. It is the fastest % among all Matlab implementations of the algorithm. % Examples % Example 1: a 5 x 5 example %{ [assignment,cost] = munkres(magic(5)); [assignedrows,dum]=find(assignment); disp(assignedrows'); % 3 2 1 5 4 disp(cost); %15 %} % Example 2: 400 x 400 random data %{ n=5; A=rand(n); tic [a,b]=munkres(A); toc %} % Reference: % "Munkres' Assignment Algorithm, Modified for Rectangular Matrices", % http://csclab.murraystate.edu/bob.pilgrim/445/munkres.html % version 1.0 by Yi Cao at Cranfield University on 17th June 2008 assignment = false(size(costMat)); cost = 0; costMat(costMat~=costMat)=Inf; validMat = costMat<Inf; validCol = any(validMat); validRow = any(validMat,2); nRows = sum(validRow); nCols = sum(validCol); n = max(nRows,nCols); if ~n return end dMat = zeros(n); dMat(1:nRows,1:nCols) = costMat(validRow,validCol); %************************************************* % Munkres' Assignment Algorithm starts here %************************************************* %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % STEP 1: Subtract the row minimum from each row. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% dMat = bsxfun(@minus, dMat, min(dMat,[],2)); %************************************************************************** % STEP 2: Find a zero of dMat. If there are no starred zeros in its % column or row start the zero. Repeat for each zero %************************************************************************** zP = ~dMat; starZ = false(n); while any(zP(:)) [r,c]=find(zP,1); starZ(r,c)=true; zP(r,:)=false; zP(:,c)=false; end while 1 %************************************************************************** % STEP 3: Cover each column with a starred zero. If all the columns are % covered then the matching is maximum %************************************************************************** primeZ = false(n); coverColumn = any(starZ); if ~any(~coverColumn) break end coverRow = false(n,1); while 1 %************************************************************************** % STEP 4: Find a noncovered zero and prime it. If there is no starred % zero in the row containing this primed zero, Go to Step 5. % Otherwise, cover this row and uncover the column containing % the starred zero. Continue in this manner until there are no % uncovered zeros left. Save the smallest uncovered value and % Go to Step 6. %************************************************************************** zP(:) = false; zP(~coverRow,~coverColumn) = ~dMat(~coverRow,~coverColumn); Step = 6; while any(any(zP(~coverRow,~coverColumn))) [uZr,uZc] = find(zP,1); primeZ(uZr,uZc) = true; stz = starZ(uZr,:); if ~any(stz) Step = 5; break; end coverRow(uZr) = true; coverColumn(stz) = false; zP(uZr,:) = false; zP(~coverRow,stz) = ~dMat(~coverRow,stz); end if Step == 6 % ************************************************************************* % STEP 6: Add the minimum uncovered value to every element of each covered % row, and subtract it from every element of each uncovered column. % Return to Step 4 without altering any stars, primes, or covered lines. %************************************************************************** M=dMat(~coverRow,~coverColumn); minval=min(min(M)); if minval==inf return end dMat(coverRow,coverColumn)=dMat(coverRow,coverColumn)+minval; dMat(~coverRow,~coverColumn)=M-minval; else break end end %************************************************************************** % STEP 5: % Construct a series of alternating primed and starred zeros as % follows: % Let Z0 represent the uncovered primed zero found in Step 4. % Let Z1 denote the starred zero in the column of Z0 (if any). % Let Z2 denote the primed zero in the row of Z1 (there will always % be one). Continue until the series terminates at a primed zero % that has no starred zero in its column. Unstar each starred % zero of the series, star each primed zero of the series, erase % all primes and uncover every line in the matrix. Return to Step 3. %************************************************************************** rowZ1 = starZ(:,uZc); starZ(uZr,uZc)=true; while any(rowZ1) starZ(rowZ1,uZc)=false; uZc = primeZ(rowZ1,:); uZr = rowZ1; rowZ1 = starZ(:,uZc); starZ(uZr,uZc)=true; end end % Cost of assignment assignment(validRow,validCol) = starZ(1:nRows,1:nCols); cost = sum(costMat(assignment));
demo_munkres.m
A=[82,83,69,92;77,37,49,92;11,69,5,86;8,9,98,23]; [assignment,cost]=munkres(A) [assignedrows,dum]=find(assignment); order=assignedrows'
3. 指派结果
>> demo_munkres assignment = 4×4 logical 数组 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 cost = 140 order = 3 2 1 4
4. 参考文献
[1] Munkres' Assignment Algorithm Modified for Rectangular Matrices
[2] Munkres Assignment Algorithm
[3] Hungarian algorithm:The assignment problem