最小生成树——Prim（普利姆）算法

【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，旨在理解Prim算法的idea 并用源代码加以实现；
0.2）最小生成树的基础知识，参见 http://blog.csdn.net/pacosonswjtu/article/details/49947085

【1】Prim算法相关

1.1）计算最小生成树的一种方法是使其连续地一步一步长成。在每一步，都要吧一个节点当做根并往上加边，这样也就把相关联的顶点加到增长中的树上；
1.2）在算法中的任一时刻，我们都可以看到一个已经添加到树上的顶点集，而其余顶点尚未加到这颗树中。此时，算法在每一阶段都可以通过选择边（u, v），使得（u, v）的值是所有u 在树上但v不在树上的边的值中的最小者，而找出一个新的顶点并吧它添加到这颗树中；
1.3）具体步骤概括为：

step1）给定一个顶点为根节点；
step2）每一步加一条边和一个顶点；（这也迎合了顶点个数-边个数=1 ）；

1.4）看个荔枝：

对上图的分析（Analysis）：
A1）可以看到，其实Prim算法基本上和求最短路径的 Dijkstra算法一样，因此和前面一样，我们对每一个顶点保留值 Dv和Pv 以及一个指标，指示该顶点是已知的还是未知的。这里，Dv是连接v 到已知顶点的最短边的权，而 Pv则是导致Dv改变的最后的顶点。
A2）算法的其余部分一样，唯一不同的是： 由于Dv的定义不同，因此它的更新法则不一样。事实上，Prim算法的更新法则比 Dijkstra算法简单：在每一个顶点ｖ被选取后，　对于每一个与　ｖ　邻接的未知的w， Dw=min（Dw, Cw,v）；

对上图的分析（Analysis）：
A1）该算法整个的实现实际上和 Dijkstra算法的实现是一样的，对于 Dijkstra算法分析所做的每一件事都可以用到这里。不过要注意， Prim算法是在无向图上运行的，因此当编写代码的时候要记住要吧每一条变都要放到两个邻接表中。
A2）不用堆时的运行时间为O(|V|^2)，它对于稠密图来说是最优的；使用二叉堆的运行时间为 O（|E|log|V|），它对于稀疏图是一个好的界限；

【2】source code + printing results（将我的代码打印结果同上图中的手动模拟的prim算法的结果进行比较，你会发现，它们的结果完全相同，这也证实了我的代码的可行性）

2.1）download source code： https://github.com/pacosonTang/dataStructure-algorithmAnalysis/tree/master/chapter9/p237_prim
2.2）source code at a glance（for complete code , please click the given link above）：

#include "prim.h"//allocate the memory for initializing unweighted table
WeightedTable *initWeightedTable(int size)
{   WeightedTable* table;int i;table = (WeightedTable*)malloc(sizeof(WeightedTable) * size);if(!table){Error("out of space ,from func initWeightedTable");return NULL;}for(i = 0; i < size; i++){table[i] = makeEmptyWeightedTable();        if(!table[i])return NULL;}return table;
} // allocate the memory for every element in unweighted table
WeightedTable makeEmptyWeightedTable()
{WeightedTable element;element = (WeightedTable)malloc(sizeof(struct WeightedTable));if(!element){Error("out of space ,from func makeEmptyWeightedTable");return NULL;}   element->known = 0; // 1 refers to accessed , also 0 refers to not accessedelement->distance = MaxInt;element->path = -1; // index starts from 0 and -1 means the startup vertex unreaches other vertexsreturn element;
}// allocate the memory for storing index of  vertex in heap and let every element -1
int *makeEmptyArray(int size)
{int *array;int i;array = (int*)malloc(size * sizeof(int));if(!array){Error("out of space ,from func makeEmptyArray");return NULL;}       for(i=0; i<size; i++)array[i] = -1;return array;
}//computing the unweighted shortest path between the vertex under initIndex and other vertexs
void prim(AdjTable* adj, int size, int startVertex, BinaryHeap bh)
{       int adjVertex;  int tempDistance;WeightedTable* table;int vertex;     AdjTable temp;  Distance tempDisStruct;int *indexOfVertexInHeap;int indexOfHeap;table = initWeightedTable(size);        tempDisStruct = makeEmptyDistance();indexOfVertexInHeap = makeEmptyArray(size);tempDisStruct->distance = table[startVertex-1]->distance;tempDisStruct->vertexIndex = startVertex-1;insert(tempDisStruct, bh, indexOfVertexInHeap); // insert the (startVertex-1) into the binary heap  table[startVertex-1]->distance = 0;// update the distance table[startVertex-1]->path = 0;// update the path of starting vertexwhile(!isEmpty(bh)){       vertex = deleteMin(bh, indexOfVertexInHeap).vertexIndex; // return the minimal element in binary heap//printBinaryHeap(bh);table[vertex]->known = 1; // update the vertex as accessed, also let responding known be 1temp = adj[vertex]->next;while(temp){adjVertex = temp->index; if(table[adjVertex]->known == 1) // judge whether table[adjVertex]->known is 1 or not{temp = temp->next;continue;}//tempDistance = table[vertex]->distance + temp->weight; // update the distancetempDistance = temp->weight;if(tempDistance < table[adjVertex]->distance){table[adjVertex]->distance = tempDistance;table[adjVertex]->path = vertex; //update the path of adjVertex, also responding path evaluated as vertex                           // key, we should judge whether adjVertex was added into the binary heap                //if true , obviously the element has been added into the binary heap(so we can't add the element into heap once again)if(indexOfVertexInHeap[adjVertex] != -1) {indexOfHeap = indexOfVertexInHeap[adjVertex];bh->elements[indexOfHeap]->distance = tempDistance; // update the distance of corresponding vertex in binary heap}else // if not ture{tempDisStruct->distance = table[adjVertex]->distance;tempDisStruct->vertexIndex = adjVertex;insert(tempDisStruct, bh, indexOfVertexInHeap); // insert the adjVertex into the binary heap}}            temp = temp->next;      }       printPrim(table, size, startVertex);        printBinaryHeap(bh);printf("\n");}       printf("\n");
} //print unweighted table
void printPrim(WeightedTable* table, int size, int startVertex)
{int i;  char *str[4] = {"vertex","known","distance","path"};printf("\n\t === storage table related to Prim alg as follows: === ");  printf("\n\t %6s%6s%9s%5s", str[0], str[1], str[2], str[3]);    for(i=0; i<size; i++){       if(i != startVertex-1 && table[i]->path!=-1) printf("\n\t %-3d   %3d   %5d      v%-3d  ", i+1, table[i]->known, table[i]->distance, table[i]->path+1);else if(table[i]->path == -1)printf("\n\t %-3d   %3d   %5d      %-3d  ", i+1, table[i]->known, table[i]->distance, table[i]->path);elseprintf("\n\t *%-3d  %3d   %5d      %-3d  ", i+1, table[i]->known, table[i]->distance, 0);}
}int main()
{ AdjTable* adj;  BinaryHeap bh;int size = 7;int capacity;int i;int j;  int startVertex;int adjTable[7][7] = {{0, 2, 4, 1, 0, 0, 0},{2, 0, 0, 3, 10, 0, 0},{4, 0, 0, 2, 0, 5, 0},{1, 3, 2, 0, 7, 8, 4},{0, 10, 0, 7, 0, 0, 6},{0, 0, 5, 8, 0, 0, 1},{0, 0, 0, 4, 6, 1, 0},};printf("\n\n\t ====== test for Prim alg finding weighted shortest path from adjoining table ======\n");adj = initAdjTable(size);       printf("\n\n\t ====== the initial weighted adjoining table is as follows:======\n");for(i = 0; i < size; i++)for(j = 0; j < size; j++)   if(adjTable[i][j])          insertAdj(adj, j, i, adjTable[i][j]); // insertAdj the adjoining table overprintAdjTable(adj, size);   capacity = 7;bh = initBinaryHeap(capacity+1);//conducting prim alg to find minimum spanning tree(MST)startVertex = 1; // you should know our index for storing vertex starts from 0prim(adj, size, startVertex, bh);   return 0;
}

2.3）printing results：