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普利姆(Prim)算法求最小生成树，也就是在包含 n个顶点的连通图中，找出只有(n-1)条边包含所有 n个顶点的连通子图，也就是所谓的极小连通子图
最小生成树：给定一个带权的无向连通图,如何选取一棵生成树,使树上所有边上权的总和为最小,这叫最小生成树
步骤：（1）从第一个点开始，找它和其它点的路径，如果不连通则设为最大，然后找出最小的路径，将此点连上
（2）如果新连入的点到某个点的路径小于原来的路径，则更新，然后继续找最小的，连上
（3）重复这几个步骤
迪杰斯特拉(Dijkstra)算法是典型最短路径算法，用于计算一个节点到其他节点的最短路径。
它的主要特点是以起始点为中心向外层层扩展(广度优先思想)，直到扩展到终点为止。
迪杰斯特拉算法和普利姆算法的区别：迪杰斯特拉是从原点到各点的路径，包括中途经过的顶点的路径也要算上，而普利姆则只算各点和它所连最短的路径那一条路径
总代码
#include<stdio.h> #include<stdlib.h> /** *建立网 */ typedef struct Net{ int
**weights; int numNodes; } *NetPtr; /** *初始化 */ NetPtr initNet(int paraSize,
int **paraData) { int i, j; NetPtr resultPtr = (NetPtr)malloc(sizeof(struct
Net)); resultPtr->numNodes = paraSize; resultPtr->weights =
(int**)malloc(sizeof(int*) * paraSize); for (i = 0; i < paraSize; i ++) {
resultPtr->weights[i] = (int*)malloc(sizeof(int) * sizeof(int)); for (j = 0; j
< paraSize; j ++) { resultPtr->weights[i][j] = paraData[i][j]; } } return
resultPtr; } /** *普利姆算法和迪杰斯特拉算法 */ int dijkstraOrPrim(NetPtr paraPtr, int
paraAlgorithm) { int i, j, minDistance, tempBestNode, resultCost; int source =
0; int numNodes = paraPtr->numNodes; int *distanceArray =
(int*)malloc(sizeof(int) * numNodes); int *parentArray =
(int*)malloc(sizeof(int) * numNodes); int *visitedArray =
(int*)malloc(sizeof(int) * numNodes); for (i = 0; i < numNodes; i ++) {
distanceArray[i] = paraPtr->weights[source][i]; parentArray[i] = source;
visitedArray[i] = 0; } distanceArray[source] = 0; parentArray[source] = -1;
visitedArray[source] = 1; tempBestNode = -1; for (i = 0; i < numNodes - 1; i
++) { minDistance = 10000; for (j = 0; j < numNodes; j ++) { if
(visitedArray[j] == 1) { continue; } if (minDistance > distanceArray[j]) {
minDistance = distanceArray[j]; tempBestNode = j; } }
visitedArray[tempBestNode] = 1; for (j = 0; j < numNodes; j ++) { if
(visitedArray[j] == 1) { continue; } if (paraPtr->weights[tempBestNode][j] >=
10000) { continue; } if (paraAlgorithm == 0) { if (distanceArray[j] >
distanceArray[tempBestNode] + paraPtr->weights[tempBestNode][j]) {
distanceArray[j] = distanceArray[tempBestNode] +
paraPtr->weights[tempBestNode][j]; parentArray[j] = tempBestNode; } } else { if
(distanceArray[j] > paraPtr->weights[tempBestNode][j]) { distanceArray[j] =
paraPtr->weights[tempBestNode][j]; parentArray[j] = tempBestNode; } } } }
printf("the parent of each node: \n"); for (i = 0; i < numNodes; i ++) {
printf("%d ", parentArray[i]); } printf("\n"); if (paraAlgorithm == 0) {
printf("From node 0, path length to all nodes are: "); for (i = 0; i <
numNodes; i ++) { printf("%d (%d), ", i, distanceArray[i]); } } else {
resultCost = 0; for (i = 0; i < numNodes; i ++) { resultCost +=
distanceArray[i]; printf("cost of node %d is %d, total = %d\n", i,
distanceArray[i], resultCost); } printf("Finally, the total cost is %d.\n",
resultCost); } printf("\n"); return resultCost; } /** *建立网 */ NetPtr
constructSampleNet() { int i, j; int myGraph[6][6] = { {0, 6, 1, 5, 0, 0}, {6,
0, 5, 0, 3, 0}, {1, 5, 0, 5, 6, 4}, {5, 0, 5, 0, 0, 2}, {0, 3, 6, 0, 0, 6}, {0,
0, 4, 2, 6, 0}}; int **tempPtr; int numNodes = 6; printf("PreParing data\n");
tempPtr = (int**)malloc(sizeof(int*) * numNodes); for (i = 0; i < numNodes; i
++) { tempPtr[i] = (int*)malloc(sizeof(int) * numNodes); } for (i = 0; i <
numNodes; i ++) { for (j = 0; j < numNodes; j ++) { if (myGraph[i][j] == 0) {
tempPtr[i][j] = 10000; } else { tempPtr[i][j] = myGraph[i][j]; } } }
printf("Data ready\n"); NetPtr resultNetPtr = initNet(numNodes, tempPtr);
return resultNetPtr; } void testPrim() { NetPtr tempNetPtr =
constructSampleNet(); printf("=====Dijkstra algorithm=====\n");
dijkstraOrPrim(tempNetPtr, 0); printf("=====Prim algorithm=====\n");
dijkstraOrPrim(tempNetPtr, 1); } int main() { testPrim(); return 0; }
运行结果
PreParing data Data ready =====Dijkstra algorithm===== the parent of each
node: -1 0 0 0 2 2 From node 0, path length to all nodes are: 0 (0), 1 (6), 2
(1), 3 (5), 4 (7), 5 (5), =====Prim algorithm===== the parent of each node: -1
2 0 5 1 2 cost of node 0 is 0, total = 0 cost of node 1 is 5, total = 5 cost of
node 2 is 1, total = 6 cost of node 3 is 2, total = 8 cost of node 4 is 3,
total = 11 cost of node 5 is 4, total = 15 Finally, the total cost is 15.