不相交集合求并的路径压缩

【0】README

0.1）本文总结于数据结构与算法分析，源代码均为原创，旨在实现对不相交集合的路径压缩操作；
0.2）对求并后的集合进行路径压缩，目的是降低集合（合并树）的深度，减少find 操作的时间复杂度；
0.3） for introduction to non-intersect set ADT ,please refet to http://blog.csdn.net/PacosonSWJTU/article/details/49716905 , and for details of finding or unionSet operations towards non-intersect set , please refer to http://blog.csdn.net/pacosonswjtu/article/details/49717009

【1】路径压缩相关

1.1）基于这样的观察：执行 Union操作的任何算法都将产生相同的最坏情形的树，因为它必然会随意打破树间的平衡。因此，无需对整个数据结构重新加工而使算法加速的唯一方法是：对Find 操作做些更聪明的工作；
1.2）路径压缩定义：设操作是Find（X），此时路径压缩的效果是，从X到根的路径上的每一个节点都使它的父节点变成根；执行find（15）后压缩路径的效果为：

对路径压缩算法的分析（Analysis）

A1）路径压缩的实施在于使用额外的两次指针移动，节点13和14 现在离根近了一个位置，而节点15和16离根近了两个位置；
A2）因此，对这些节点未来的快速访问将由于花费额外的工作来进行路径压缩而得到补偿；

1.3）路径压缩对基本的 Find操作改变不大。对 Find 操作来说，唯一的变化是使得 S[X] 等于由Find 返回的值；这样，在集合的根被递归地找到以后， X 就直接指向了它，对通向根的路径上的每一个节点这将递归地出现，因此实现了路径压缩；
1.4）路径压缩可以和大小求并完全兼容，这就使得两个例程可以同时实现；
1.5）路径压缩不完全与按高度求并兼容，因为路径压缩可以改变树的高度；

【2】source code + printing results

2.1）download source code：
https://github.com/pacosonTang/dataStructure-algorithmAnalysis/blob/master/chapter8/p206_pathCompression.c

souce code statements：路径压缩源代码中使用的集合求并方法是： 按大小求并：

2.2）source code at a glance：

#include <stdio.h>
#include <malloc.h>#define ElementType int
#define Error(str) printf("\n error: %s \n",str) struct UnionSet;
typedef struct UnionSet* UnionSet;// we adopt the child-sibling expr
struct UnionSet
{int parent;int size;ElementType value;
};UnionSet makeEmpty();
UnionSet* initUnionSet(int size, ElementType* data);
void printSet(UnionSet* set, int size);
void printArray(ElementType data[], int size);
int find(int index, UnionSet* set);
void pathCompress(int, UnionSet*);// initialize the union set
UnionSet* initUnionSet(int size, ElementType* data)
{UnionSet* set;  int i;set = (UnionSet*)malloc(size * sizeof(UnionSet));if(!set){Error("out of space, from func initUnionSet");        return NULL;}   for(i=0; i<size; i++){set[i] = makeEmpty();if(!set[i])return NULL;set[i]->value = data[i];}return set;
}// allocate the memory for the single UnionSet and evaluate the parent and size -1
UnionSet makeEmpty()
{UnionSet temp;temp = (UnionSet)malloc(sizeof(struct UnionSet));if(!temp){Error("out of space, from func makeEmpty!");        return NULL;}temp->parent = -1;temp->size = 1;return temp;
}// merge set1 and set2 by size
void setUnion(UnionSet* set, int index1, int index2)
{//judge whether the index1 or index2 equals to -1 ,also -1 represents the rootif(index1 != -1)index1 = find(index1, set);if(index2 != -1)index2 = find(index2, set);if(set[index1]->size > set[index2]->size){set[index2]->parent = index1;set[index1]->size += set[index2]->size;}else{set[index1]->parent = index2;set[index2]->size += set[index1]->size;}
} //find the root of one set whose value equals to given value
int find(int index, UnionSet* set)
{UnionSet temp;  while(1){temp = set[index];if(temp->parent == -1)break;index = temp->parent;}return index;
}   // conducting path compression towards union set with given index
void pathCompress(int index, UnionSet* set)
{   int root;int i;int parent;//1st step: find the top root contains the element under index  root = find(index, set);//2nd step: path compression beginsi = set[index]->parent;set[index]->parent = root;while(i != -1) {       parent = set[i]->parent;                if(parent == root)break;set[i]->parent = root;i = parent;}
}int main()
{int size;UnionSet* unionSet;ElementType data[] = {110, 245, 895, 658, 321, 852, 147, 458, 469, 159, 347, 28};size = 12;printf("\n\t====== test for conducting path compression towards union set by size ======\n");//printf("\n\t=== the initial array is as follows ===\n");//printArray(data, depth); printf("\n\t=== the init union set are as follows ===\n");unionSet = initUnionSet(size, data); // initialize the union set over//printSet(unionSet, size);printf("\n\t=== after union(0, 1) + union(2, 3) + union(4, 5) + union(6, 7) + union(8, 9) + union(10 ,11) ===\n");setUnion(unionSet, 0, 1);setUnion(unionSet, 2, 3);setUnion(unionSet, 4, 5);setUnion(unionSet, 6, 7);setUnion(unionSet, 8, 9);   setUnion(unionSet, 10, 11); //printSet(unionSet, size);printf("\n\t=== after union(1, 3) + union(5, 7) + union(9, 11) ===\n");setUnion(unionSet, 1, 3);setUnion(unionSet, 5, 7);setUnion(unionSet, 9, 11);//printSet(unionSet, size);  printf("\n\t=== after union(3, 7) + union(7, 11) ===\n");setUnion(unionSet, 3, 7);setUnion(unionSet, 7, 11);  printSet(unionSet, size); printf("\n\t=== after pathCompress(0, unionSet) + pathCompress(8, unionSet) ===\n");pathCompress(0, unionSet) ;pathCompress(8, unionSet);printSet(unionSet, size); return 0;
}void printArray(ElementType data[], int size)
{int i;for(i = 0; i < size; i++)    printf("\n\t data[%d] = %d", i, data[i]);                    printf("\n\n");
} void printSet(UnionSet* set, int size)
{int i;UnionSet temp;for(i = 0; i < size; i++){       temp = set[i];printf("\n\t parent[%d] = %d", i, temp->parent);                }printf("\n");
}

2.3）printing results：