# Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)

• Difficulty Level : Hard
• Last Updated : 20 Jul, 2022

1. For each row of the matrix, find the smallest element and subtract it from every element in its row.
2. Do the same (as step 1) for all columns.
3. Cover all zeros in the matrix using minimum number of horizontal and vertical lines.
4. Test for Optimality: If the minimum number of covering lines is n, an optimal assignment is possible and we are finished. Else if lines are lesser than n, we havenâ€™t found the optimal assignment, and must proceed to step 5.
5. Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.

Try it before moving to see the solution

Explanation for above simple example:

```
Below is the cost matrix of example given in above diagrams.
2500  4000  3500
4000  6000  3500
2000  4000  2500

Step 1: Subtract minimum of every row.
2500, 3500 and 2000 are subtracted from rows 1, 2 and
3 respectively.

0   1500  1000
500  2500   0
0   2000  500

Step 2: Subtract minimum of every column.
0, 1500 and 0 are subtracted from columns 1, 2 and 3
respectively.

0    0   1000
500  1000   0
0   500  500

Step 3: Cover all zeroes with minimum number of
horizontal and vertical lines.

Step 4:  Since we need 3 lines to cover all zeroes,
we have found the optimal assignment.
2500  4000  3500
4000  6000  3500
2000  4000  2500

So the optimal cost is 4000 + 3500 + 2000 = 9500```

An example that doesn’t lead to optimal value in first attempt: In the above example, the first check for optimality did give us solution. What if we the number covering lines is less than n.

```
cost matrix:
1500  4000  4500
2000  6000  3500
2000  4000  2500

Step 1: Subtract minimum of every row.
1500, 2000 and 2000 are subtracted from rows 1, 2 and
3 respectively.

0    2500  3000
0    4000  1500
0    2000   500

Step 2: Subtract minimum of every column.
0, 2000 and 500 are subtracted from columns 1, 2 and 3
respectively.

0     500  2500
0    2000  1000
0      0      0

Step 3: Cover all zeroes with minimum number of
horizontal and vertical lines.

Step 4:  Since we only need 2 lines to cover all zeroes,

Step 5:  We subtract the smallest uncovered entry
from all uncovered rows. Smallest entry is 500.
-500    0   2000
-500  1500   500
0     0      0

Then we add the smallest entry to all covered columns, we get
0     0   2000
0   1500   500
500    0      0

lines. and go to Step 4:. Since we need 3 lines to
cover, we found the optimal solution.
1500  4000  4500
2000  6000  3500
2000  4000  2500

So the optimal cost is 4000 + 2000 + 2500 = 8500```

## C++

 `#include``using` `namespace` `std;`  `class` `Solution {``  ``public``:``  ` `    ``int` `cost[31][31]; ``//cost matrix``    ``int` `n, max_match; ``//n workers and n jobs``    ``int` `lx[31], ly[31]; ``//labels of X and Y parts``    ``int` `xy[31]; ``//xy[x] - vertex that is matched with x,``    ``int` `yx[31]; ``//yx[y] - vertex that is matched with y``    ``bool` `S[31], T[31]; ``//sets S and T in algorithm``    ``int` `slack[31]; ``//as in the algorithm description``    ``int` `slackx[31]; ``//slackx[y] such a vertex, that``    ``int` `prev_ious[31]; ``//array for memorizing alternating p``  ` `    ``void` `init_labels()``    ``{``        ``memset``(lx, 0, ``sizeof``(lx));``        ``memset``(ly, 0, ``sizeof``(ly));``        ``for` `(``int` `x = 0; x < n; x++)``        ``for` `(``int` `y = 0; y < n; y++)``        ``lx[x] = max(lx[x], cost[x][y]);``    ``}``    ` `     ` `    ``void` `update_labels()``    ``{``        ``int` `x, y;``        ``int` `delta = 99999999; ``//init delta as infinity``        ``for` `(y = 0; y < n; y++) ``//calculate delta using slack``            ``if` `(!T[y])``                ``delta = min(delta, slack[y]);``        ``for` `(x = 0; x < n; x++) ``//update X labels``            ``if` `(S[x])``                ``lx[x] -= delta;``        ``for` `(y = 0; y < n; y++) ``//update Y labels``            ``if` `(T[y])``                ``ly[y] += delta;``        ``for` `(y = 0; y < n; y++) ``//update slack array``            ``if` `(!T[y])``                ``slack[y] -= delta;``    ``}``    ` `    ` `    ``void` `add_to_tree(``int` `x, ``int` `prev_iousx)``    ``//x - current vertex,prev_iousx - vertex from X before x in the alternating path,``    ``//so we add edges (prev_iousx, xy[x]), (xy[x], x)``    ``{``        ``S[x] = ``true``; ``//add x to S``        ``prev_ious[x] = prev_iousx; ``//we need this when augmenting``        ``for` `(``int` `y = 0; y < n; y++) ``//update slacks, because we add new vertex to S``            ``if` `(lx[x] + ly[y] - cost[x][y] < slack[y])``            ``{``                ``slack[y] = lx[x] + ly[y] - cost[x][y];``                ``slackx[y] = x;``            ``}``    ``}``    ` `    ` `    ` `    ``void` `augment() ``//main function of the algorithm``    ``{``        ``if` `(max_match == n) ``return``; ``//check whether matching is already perfect``        ``int` `x, y, root; ``//just counters and root vertex``        ``int` `q[31], wr = 0, rd = 0; ``//q - queue for bfs, wr,rd - write and read``        ``//pos in queue``        ``memset``(S, ``false``, ``sizeof``(S)); ``//init set S``        ``memset``(T, ``false``, ``sizeof``(T)); ``//init set T``        ``memset``(prev_ious, -1, ``sizeof``(prev_ious)); ``//init set prev_ious - for the alternating tree``        ` `        ``for` `(x = 0; x < n; x++) ``//finding root of the tree``        ``{``            ``if` `(xy[x] == -1)``            ``{``                ``q[wr++] = root = x;``                ``prev_ious[x] = -2;``                ``S[x] = ``true``;``                ``break``;``            ``}``        ``}``        ` `        ``for` `(y = 0; y < n; y++) ``//initializing slack array``        ``{``            ``slack[y] = lx[root] + ly[y] - cost[root][y];``            ``slackx[y] = root;``        ``}``        ` `        ``//second part of augment() function``        ``while` `(``true``) ``//main cycle``        ``{``            ``while` `(rd < wr) ``//building tree with bfs cycle``            ``{``                ``x = q[rd++]; ``//current vertex from X part``                ``for` `(y = 0; y < n; y++) ``//iterate through all edges in equality graph``                    ``if` `(cost[x][y] == lx[x] + ly[y] && !T[y])``                    ``{``                        ``if` `(yx[y] == -1) ``break``; ``//an exposed vertex in Y found, so``                                                ``//augmenting path exists!``                            ``T[y] = ``true``; ``//else just add y to T,``                        ``q[wr++] = yx[y]; ``//add vertex yx[y], which is matched``                        ``//with y, to the queue``                        ``add_to_tree(yx[y], x); ``//add edges (x,y) and (y,yx[y]) to the tree``                    ``}``                ``if` `(y < n)``                    ``break``; ``//augmenting path found!``            ``}``            ``if` `(y < n)``                ``break``; ``//augmenting path found!``            ` `            ``update_labels(); ``//augmenting path not found, so improve labeling``            ` `            ``wr = rd = 0;``            ``for` `(y = 0; y < n; y++)``            ``//in this cycle we add edges that were added to the equality graph as a``            ``//result of improving the labeling, we add edge (slackx[y], y) to the tree if``            ``//and only if !T[y] && slack[y] == 0, also with this edge we add another one``            ``//(y, yx[y]) or augment the matching, if y was exposed``            ``if` `(!T[y] && slack[y] == 0)``            ``{``                ``if` `(yx[y] == -1) ``//exposed vertex in Y found - augmenting path exists!``                ``{``                    ``x = slackx[y];``                    ``break``;``                ``}``                ``else``                ``{``                    ``T[y] = ``true``; ``//else just add y to T,``                    ``if` `(!S[yx[y]])``                    ``{``                        ``q[wr++] = yx[y]; ``//add vertex yx[y], which is matched with``                        ``//y, to the queue``                        ``add_to_tree(yx[y], slackx[y]); ``//and add edges (x,y) and (y,``                        ``//yx[y]) to the tree``                    ``}``                ``}``            ``}``            ``if` `(y < n) ``break``; ``//augmenting path found!``        ``}``        ` `        ``if` `(y < n) ``//we found augmenting path!``        ``{``            ``max_match++; ``//increment matching``            ``//in this cycle we inverse edges along augmenting path``            ``for` `(``int` `cx = x, cy = y, ty; cx != -2; cx = prev_ious[cx], cy = ty)``            ``{``                ``ty = xy[cx];``                ``yx[cy] = cx;``                ``xy[cx] = cy;``            ``}``            ``augment(); ``//recall function, go to step 1 of the algorithm``        ``}``    ``}``//end of augment() function``     ` `    ``int` `hungarian()``    ``{``        ``int` `ret = 0; ``//weight of the optimal matching``        ``max_match = 0; ``//number of vertices in current matching``        ``memset``(xy, -1, ``sizeof``(xy));``        ``memset``(yx, -1, ``sizeof``(yx));``        ``init_labels(); ``//step 0``        ``augment(); ``//steps 1-3``        ` `        ``for` `(``int` `x = 0; x < n; x++) ``//forming answer there``            ``ret += cost[x][xy[x]];``    ` `        ``return` `ret;``    ``}``    ` `    ``int` `assignmentProblem(``int` `Arr[], ``int` `N) {``        ` `        ``n = N;``        ``for``(``int` `i=0; i

Output

`8500`