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最小生成树——Prim(普利姆)算法

發布時間:2023/12/3 编程问答 31 豆豆
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【0】README

0.1) 本文總結于 數據結構與算法分析, 源代碼均為原創, 旨在 理解Prim算法的idea 并用 源代碼加以實現;
0.2)最小生成樹的基礎知識,參見 http://blog.csdn.net/pacosonswjtu/article/details/49947085


【1】Prim算法相關

1.1)計算最小生成樹的一種方法是使其連續地一步一步長成。在每一步, 都要吧一個節點當做根并往上加邊,這樣也就把相關聯的頂點加到增長中的樹上;
1.2)在算法中的任一時刻, 我們都可以看到一個已經添加到樹上的頂點集, 而其余頂點尚未加到這顆樹中。此時, 算法在每一階段都可以通過選擇邊(u, v),使得(u, v)的值是所有u 在樹上但v不在樹上的邊的值中的最小者, 而找出一個新的頂點并吧它添加到這顆樹中;
1.3)具體步驟概括為:

  • step1)給定一個頂點為根節點;
  • step2)每一步加一條邊和一個頂點; (這也迎合了 頂點個數-邊個數=1 );

1.4)看個荔枝:

對上圖的分析(Analysis):
A1)可以看到, 其實Prim算法基本上和求最短路徑的 Dijkstra算法一樣, 因此和前面一樣,我們對每一個頂點保留值 Dv和Pv 以及一個指標,指示該頂點是已知的還是未知的。這里,Dv是連接v 到已知頂點的最短邊的權, 而 Pv則是導致Dv改變的最后的頂點。
A2)算法的其余部分一樣, 唯一不同的是: 由于Dv的定義不同, 因此它的更新法則不一樣。事實上,Prim算法的更新法則比 Dijkstra算法簡單:在每一個頂點v被選取后, 對于每一個與 v 鄰接的未知的w, Dw=min(Dw, Cw,v);

對上圖的分析(Analysis):
A1)該算法整個的實現實際上和 Dijkstra算法的實現是一樣的, 對于 Dijkstra算法分析所做的每一件事都可以用到這里。 不過要注意, Prim算法是在無向圖上運行的, 因此當編寫代碼的時候要記住要吧每一條變都要放到兩個鄰接表中。
A2)不用堆時的運行時間為O(|V|^2), 它對于稠密圖來說是最優的; 使用二叉堆的運行時間為 O(|E|log|V|), 它對于稀疏圖是一個好的界限;


【2】source code + printing results(將我的代碼打印結果 同 上圖中的手動模擬的prim算法的結果進行比較,你會發現, 它們的結果完全相同,這也證實了我的代碼的可行性)

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:



總結

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