最大生成树算法C语言

  只是将算法的主例程边的权值改为异号的,即可完成这个算法.我的初衷是将最小堆改成最大堆.但事实证明,这是行不通的.贴代码.

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/* 9-20-12-22-11.16.c -- 第九章第二十题 */ #include #include #include "binary_heap_for_kruskal.h" #include "disjiont_set.h" #define SIZE (10) int main (void) ; void kruskal (const Adjacenty_List * const padj, const Hash_Table * const pht, const int start, char (* const result)[2], int * const weight) ; void print_result (const char (* const result)[2], const int * const weight, const int size) ; int main (void) { Adjacenty_List adj ; Hash_Table ht ; int capacity = 10 ; char result[SIZE][2] ; int weight[SIZE], size = SIZE ; Initialize_H (&ht, capacity * 2) ; Initialize_A (&adj, capacity) ; /* Initialize the adjacenty list */ InitializeALine_A (&adj, &ht, 0, 'a', 0, 6, 'd', 4, 'e', 4, 'b', 3) ; InitializeALine_A (&adj, &ht, 1, 'b', 0, 8, 'a', 3, 'e', 2, 'f', 3, 'c', 10) ; InitializeALine_A (&adj, &ht, 2, 'c', 0, 6, 'b', 10, 'f', 6, 'g', 1) ; InitializeALine_A (&adj, &ht, 3, 'd', 0, 6, 'a', 4, 'e', 5, 'h', 6) ; InitializeALine_A (&adj, &ht, 4, 'e', 0, 12, 'd', 5, 'a', 4, 'b', 2, 'f', 11, 'i', 1, 'h', 2) ; InitializeALine_A (&adj, &ht, 5, 'f', 0, 12, 'e', 11, 'b', 3, 'c', 6, 'g', 2, 'j', 11, 'i', 3) ; InitializeALine_A (&adj, &ht, 6, 'g', 0, 6, 'f', 2, 'c', 1, 'j', 8) ; InitializeALine_A (&adj, &ht, 7, 'h', 0, 6, 'd', 6, 'e', 2, 'i', 4) ; InitializeALine_A (&adj, &ht, 8, 'i', 0, 8, 'h', 4, 'e', 1, 'f', 3, 'j', 7) ; InitializeALine_A (&adj, &ht, 9, 'j', 0, 6, 'i', 7, 'f', 11, 'g', 8) ; kruskal (&adj, &ht, 0, result, weight) ; print_result (result, weight, size - 1) ; Release_H (&ht) ; Release_A (&adj) ; return 0 ; } void kruskal (const Adjacenty_List * const padj, const Hash_Table * const pht, const int start, char (* const result)[2], int * const weight) { Adjoin_To_Vertex * scan ; Disjiont d ; SetType vset, wset ; Binary_Heap bh ; Edge edge ; int capacity = (*padj) -> capacity, vertex_sub = capacity, i ; d = (int *) malloc (sizeof (int) * ((*pht) -> capacity + 1)) ; if (NULL == d) return ; /* Make sure the size of Disjiont is enough large. */ Initialize_D (d, (*pht) -> capacity) ; /* Every edge has two vertexs and every edge will enqueueing twice. */ Initialize_B (&bh, capacity * 4) ; for (i = 0; i < capacity; i++) { edge.v_hash_value = (*padj) -> list[i].hash_value ; scan = (*padj) -> list[i].adjoin_to ; while (scan) { edge.w_hash_value = scan -> hash_value ; /* This is the crux of the matter */ edge.weight = -scan -> cvw ; Insert_B (&bh, &edge) ; scan = scan -> next ; } } i = 0 ; while (i < vertex_sub - 1) { DeleteMin_B (&bh, &edge) ; vset = Find_D (d, edge.v_hash_value) ; wset = Find_D (d, edge.w_hash_value) ; if (vset != wset) { Union_D (d, vset, wset) ; result[i][0] = (*pht) -> lists[edge.v_hash_value].name ; result[i][1] = (*pht) -> lists[edge.w_hash_value].name ; weight[i] = edge.weight ; i++ ; } } free (d) ; Release_B (&bh) ; } void print_result (const char (* const result)[2], const int * const weight, const int size) { int i ; for (i = 0; i < size; i++) printf ("%c to %c is %d/n", result[i][0], result[i][1], weight[i]) ; }