最小生成树Kruskal算法C语言

/* 9-18(b)-12-19-00.49.c -- 第九章第十八题 */ #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 ; 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]) ; }   这个东西,写了4小时多.

  思路比较清晰,主要的过程是在讲实际问题模拟成代码.这次进行了计划,按部就班.于是,变化就比较少,完成得就比较快.欲速则不达是必须的.

  白天干活比较累,一种让人绝望的感觉.加油吧.抓紧时间,把时间抓射,呵呵.

  总之啊,每天都在学习.即使是白天从事毫不相关的工作,我也在无时无刻地学习,各种各种了,让我更智慧,更具竞争力.

  还有,我回忆了下不相交集合.一个不相交集合是很多不相交集合的集合,这很重要.因为这段代码,我加深了理解.

  自己也写了新的堆.这次的堆写得比较顺畅了.

  还有,就是传递多维数组.都已经忘了,看了下C语言的书,想起来了.温故而知新,真的很重要,不然真就完全剩下思想了.

  我想说的是,我学得越扎实,越好,越快.也就会更早地摆脱当前的境地.加油吧.

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