Венгерский алгоритм - C#
Формулировка задачи:
Ребята прошу Вашей помощи, нужен исходник Венгерского алгоритма на C#
Решение задачи: «Венгерский алгоритм»
textual
Листинг программы
using System; public class OptimalAssignmentSolver { /// <summary> /// Solves optimal assignment problem using the Kuhn-Munkres /// (aka Hungarian) algorithm. Time complexity: O(n^3). /// /// References: /// [url]http://www.math.uwo.ca/~mdawes/courses/344/kuhn-munkres.pdf[/url] /// [url]http://www.ecp6.jussieu.fr/pageperso/bondy/books/gtwa/pdf/chapter5.pdf[/url] /// </summary> /// <param name="a">A square matrix.</param> /// <returns> /// A permutation p of integers 0, 1, ..., n-1, (where /// n is the size of matrix a) such that the expression /// a[0, p[0]] + a[1, p[1]] + ... + a[n-1, p[n-1]] /// is maximum possible among all such permutations. /// </returns> public static int[] KuhnMunkres(int[,] a) { int N = a.GetLength(0); if (N == 0) return new int[0]; int[] lx = new int[N], ly = new int[N]; // labelling function for vertices in first and second partitions int[] mx = new int[N], my = new int[N]; // mx[u]=v, my[v]=u <==> u and v are currently matched; -1 values means 'not matched' int[] px = new int[N], py = new int[N]; // predecessor arrays. used in DFS to reconstruct paths. int[] stack = new int[N]; // invariant: lx[u] + ly[v] >= a[u, v] // (implies that any perfect matching in subgraph containing only // edges u, v for which lx[u]+ly[v]=a[u,v] is the optimal matching.) // compute initial labelling function: lx[i] = max_j(a[i, j]), ly[j] = 0; for (int i = 0; i < N; i++) { lx[i] = a[i, 0]; for (int j = 0; j < N; j++) if (a[i, j] > lx[i]) lx[i] = a[i, j]; ly[i] = 0; mx[i] = my[i] = -1; } for (int size = 0; size < N;) { int s; for (s = 0; mx[s] != -1; s++); // s is an unmatched vertex in the first partition. // At the current iteration we will either find an augmenting path // starting at s, or we'll extend the equality subgraph so that // such a path will exist at the next iteration. for (int i = 0; i < N; i++) px[i] = py[i] = -1; px[s] = s; // DFS int t = -1; stack[0] = s; for (int top = 1; top > 0;) { int u = stack[--top]; for (int v = 0; v < N; v++) { if (lx[u] + ly[v] == a[u, v] && py[v] == -1) { if (my[v] == -1) { // we've found an augmenting path t = v; py[t] = u; top = 0; break; } py[v] = u; px[my[v]] = v; stack[top++] = my[v]; } } } if (t != -1) { // augment along the found path while (true) { int u = py[t]; mx[u] = t; my[t] = u; if (u == s) break; t = px[u]; } ++size; } else { // No augmenting path exists from s in the current equality graph, // Modify labelling function a bit... int delta = int.MaxValue; for (int u = 0; u < N; u++) { if (px[u] == -1) continue; for (int v = 0; v < N; v++) { if (py[v] != -1) continue; int z = lx[u] + ly[v] - a[u, v]; if (z < delta) delta = z; } } for (int i = 0; i < N; i++) { if (px[i] != -1) lx[i] -= delta; if (py[i] != -1) ly[i] += delta; } } } // Verify optimality bool correct = true; for (int u = 0; u < N; u++) { for (int v = 0; v < N; v++) { correct &= (lx[u] + ly[v] >= a[u, v]); if (mx[u] == v) correct &= (lx[u] + ly[v] == a[u, v]); if (!correct) { throw new Exception( "*** Internal error: optimality conditions are not satisfied ***\n" + "Most probably an overflow occurred"); } } } return mx; } }
ИИ поможет Вам:
- решить любую задачу по программированию
- объяснить код
- расставить комментарии в коде
- и т.д