不相交集的求并算法（按集合大小求并+按高度求并）

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【0】README

0.1）本文总结于数据结构与算法分析，但源代码均为原创，旨在实现不相交集ADT的两个操作：合并集合union+查找集合find；
0.2） 不相交集ADT 的 Introduction ，参见 http://blog.csdn.net/PacosonSWJTU/article/details/49716905

【1】灵巧求并算法——按集合大小求并

1.1）大小求并法定义：上面的Union执行是相当任意的，通过使第二棵树成为第一棵树的子树而完成合并；对其的改进是借助任意的方法打破现有关系，使得总让较小的树成为较大树的子树，我们把这种方法叫做 大小求并法；
1.2）可以看到， 如果Union 操作都是按照大小求并的话，那么任何节点的深度均不会超过 logN；
1.3）首先注意节点的深度为0，然后它的深度随着一次 Union 的结果而增加的时候，该节点则被置于至少是它以前所在树两倍大的一棵树上；因此，它的深度最多可以增加 logN次；

1.3.2） Find 操作的运行时间为 O（logN），而连续M次操作则花费 O（MlogN）；
1.3.3） 下图指出在16次Union操作后可能得到这种最坏的树；而且如果所有的Union都对相等大小的树进行，那么这样的树是会得到的；

1.4）为了实现这种方法，我们需要记住每个树的大小。由于我们实际上只使用一个数组，因此可以让每个根的数组元素包含它的树的大小的负值；
1.5）已经证明：若使用按大小求并则连续 M次运算需要 O（M）平均时间，这是因为当随机的Union执行时，整个算法一般只有一些很小的集合（通常是一个元素）与大集合合并；

1.6）souce code + printing

1.6.1）download source code ：
https://github.com/pacosonTang/dataStructure-algorithmAnalysis/blob/master/chapter8/p203_unionBySize.c
Source Code Statements：
- S1）显然，本源代码只对元素的size进行了加，没有减；因为我想的话，元素只能合并一次，也即是元素C起初合并到了集合A，就不能再次合并到集合B，元素C就一直属于集合A的子集了；
- S2）当然，这个代码只是大致上实现了按大小求并的思想，如果元素C还可以再次合并到其他集合的话，这就涉及到集合根元素的size的加减问题了；需要的话，添加之即可；
1.6.2）souce 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 exprstruct 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(ElementType index, UnionSet* set);// 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 -1UnionSet 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 sizevoid setUnion(UnionSet* set, int index1, int index2){    //judge whether the index1 or index2 equals to -1 ,also -1 represents the root    if(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 valueint find(ElementType index, UnionSet* set) {    UnionSet temp;      while(1)    {        temp = set[index];        if(temp->parent == -1)            break;        index = temp->parent;    }    return index;       }   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 union set by size ======\n");    //printf("\n\t=== the initial array is as follows ===\n");    //printArray(data, size);     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(1,5) + union(2,5) + union(3,4) + union(4,5) ===\n");    setUnion(unionSet, 1, 5);    setUnion(unionSet, 2, 5);    setUnion(unionSet, 3, 4);    setUnion(unionSet, 4, 5);    printSet(unionSet, size);    printf("\n\t=== after union(9,8) + union(7,6) + union(3,6) ===\n");    setUnion(unionSet, 9, 8);    setUnion(unionSet, 7, 6);    setUnion(unionSet, 3, 6);       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");}

1.6.3）printing result

【2】灵巧求并算法——按集合高度求并

2.1）按高度求并定义： 另外一种方法是按照高度求并（按照高度求并是按大小求并的简单修改）（推荐）；
2.2）它同样保证所有的树的深入最多是 O（logN)。我们使得浅的树成为深的树的子树，这是一种平缓算法， 因为只有当两颗相等深度的树求并时树的高度才会增加（此时树的高度增加1）。
2.3）source code + printing result

2.3.1）download source code ：
https://github.com/pacosonTang/dataStructure-algorithmAnalysis/blob/master/chapter8/p205_unionByHeight.c
2.3.2）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 depth-sibling exprstruct UnionSet{    int parent;    int height;    ElementType value;};UnionSet makeEmpty();UnionSet* initUnionSet(int depth, ElementType* data);void printSet(UnionSet* set, int depth);void printArray(ElementType data[], int depth);int find(ElementType index, UnionSet* set);// 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 depth -1UnionSet makeEmpty(){    UnionSet temp;    temp = (UnionSet)malloc(sizeof(struct UnionSet));    if(!temp)    {        Error("out of space, from func makeEmpty!");                return NULL;    }    temp->parent = -1;    temp->height = 0;    return temp;}// merge set1 and set2 by depthvoid setUnion(UnionSet* set, int index1, int index2){    //judge whether the index1 or index2 equals to -1 ,also -1 represents the root    if(index1 != -1)        index1 = find(index1, set);    if(index2 != -1)        index2 = find(index2, set);    if(set[index1]->height > set[index2]->height)           set[index2]->parent = index1;               else if(set[index1]->height < set[index2]->height)          set[index1]->parent = index2;           else    {        set[index1]->parent = index2;        set[index2]->height++; // only if the height of both of subtrees is equal, the height increases one    }} //find the root of one set whose value equals to given valueint find(ElementType index, UnionSet* set) {    UnionSet temp;      while(1)    {        temp = set[index];        if(temp->parent == -1)            break;        index = temp->parent;    }    return index;       }   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 union set by depth ======\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);     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");}

2.3.3）printing results：

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